Real Analysis: Series
Convergence Tests
Motivation
Determining whether a given series converges or diverges in bespoke fashion can be tedious. In order to hasten the process, there are several theorems about convergence we can prove that can make such determinations much easier. Each of these theorems allows us to use algebraic manipulations to compare individual terms of the underlying sequence either to each other or to a more easily analyzed sequences in simpler series. The application of these theorems to determine the convergence or divergence of particular series is its own area of calculus.
Note that all of these tests only provide information about whether a series converges. Determining whether the limit of a convergent series can be expressed in simple terms such as a fraction or algebraic expression is another matter entirely.
The Absolute Convergence Test
A sum $\sum\limits_{n=1}^{\infty} a_n$ converges absolutely (or is absolutely convergent) if $\sum\limits_{n=1}^{\infty} |a_n|$ converges. The absolute convergence test states that a series that converges absolutely also converges. How handy! However, the converse is not necessarily true - a convergent series is not always absolutely convergent.
The Comparison Test
The comparison test comes in two parts, one for convergence and one for divergence. Namely, if we are investigating whether a series $\sum\limits_{n=1}^{\infty} a_n$ converges, we ask for another series, $\sum\limits_{n=1}^{\infty} b_n$, that we know surely converges or diverges. We then compare the individual $a_n$ and $b_n$ terms.
Convergence: If $0 \leq a_n \leq b_n$ for all $n \in \mathbb{N}^+,$ and $\sum\limits_{n=1}^{\infty} b_n$ converges, then $\sum\limits_{n=1}^{\infty} a_n$ also converges.
Divergence: If $0 \leq a_n \geq b_n$ for all $n \in \mathbb{N}^+,$ and $\sum\limits_{n=1}^{\infty} b_n$ diverges, then $\sum\limits_{n=1}^{\infty} a_n$ also diverges.
The comparison test can be useful for quickly determining whether a series converges or diverges if we have one with known convergence characteristics on hand. Furthermore, its simplicity makes it one of the most important convergences tests in the list, especially as it is used in the proofs of several others. However, a drawback of the comparison test is precisely that it requires a secondary series; we cannot use the comparison test to analyze the original series itself in isolation.
The Comparison Limit Test
Closely related to the comparison test is the comparison limit test, which states the following: if $\{a_n\}$ and $\{b_n\}$ are two sequences where $a_n \geq 0$ and $b_n > 0$ for all $n \in \mathbb{N}^+$ and $0 < \lim\limits_{n \rightarrow \infty} \frac{a_n}{b_n} < \infty,$ then either $\sum\limits_{n=1}^{\infty} a_n$ and $\sum\limits_{n=1}^{\infty} b_n$ either both converge or both diverge.
The Cauchy Condensation Test
Consider the series $\sum\limits_{n=1}^{\infty} a_n,$ where $\{a_n\}$ is positive and monotonically decreasing. The Cauchy condensation test states that $\sum\limits_{n=1}^{\infty} a_n$ converges if and only if $\sum\limits_{n=1}^{\infty} 2^na_{2^n}$ converges. The term "condensation" in the name of the test comes from the fact that only a condensed number of terms in the sequence $\{a_n\}$ need to be analyzed, rather than all of them.
The Cauchy condensation test can be used to more easily prove that the harmonic series diverges. In fact, the proof of this test is very similar to the proof of the harmonic series's divergence. Furthermore, the Cauchy condensation tests can be used to show the convergence of the generalized harmonic series under the right conditions.
The Ratio Test
Unlike the comparison test, the ratio test does not need an additional series to work with, as it only makes use of the terms in the series' underlying sequence. Furthermore, the ratio test comes in three parts, one each for convergence and divergence, as well as a third that is inconclusive. When used to determine whether a series $\sum\limits_{n=1}^{\infty} a_n$, the ratio test considers the limit of sequence $\left\{\dfrac{a_{n+1}}{a_n}\right\},$ which can fall into one of three cases:
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Convergence: If $\lim\limits_{n \rightarrow \infty} \left|\frac{a_{n+1}}{a_n}\right| < 1,$ then $\sum\limits_{n=1}^{\infty} a_n$ converges.
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Divergence: If $\lim\limits_{n \rightarrow \infty} \left|\frac{a_{n+1}}{a_n}\right| > 1$ or $\lim\limits_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} = \infty,$ then $\sum\limits_{n=1}^{\infty}$ diverges.
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Inconclusive: If $\lim\limits_{n \rightarrow \infty} \left|\frac{a_{n+1}}{a_n}\right| =1,$ then the ratio test is inconclusive, and some other form of analysis is needed to determine whether $\sum\limits_{n=1}^{\infty} a_n$ converges or diverges.
The ratio test can be generalized to use the limit infimum and limit supremum to cover cases where the ratio's limit does not exist but its limit superior or limit inferior exist:
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Convergence: If $\limsup\limits_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1,$ then $\sum\limits_{n=1}^{\infty} a_n$ converges absolutely.
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Divergence: If $\liminf\limits_{n \rightarrow \infty} \left| \frac{a_{n+1}}{a_n} \right| > 1,$ then $\sum\limits_{n=1}^{\infty} a_n$ converges absolutely.
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Divergence: If there exists an $N \in \mathbb{N}$ such that $\left| \frac{a_{n+1}}{a_n} \right| > 1$ whenever $n > N,$ then $\sum\limits_{n=1}^{\infty} a_n$ diverges.
The ratio test for convergence suffers from the same problem as the comparison test for convergence in that it also does not give any information as to what value the series actually converges to.
The Root Test
The root test only requires the original sequence being summed in the series. Given a series $\sum\limits_{n=1}^{\infty} a_n,$ the root test states the following:
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Convergence: If $\lim\limits_{n \rightarrow \infty} |a_n|^{\frac{1}{n}} < 1,$ then $\sum\limits_{n=1}^{\infty} a_n$ converges.
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Divergence: If $\lim\limits_{n \rightarrow \infty} |a_n|^{\frac{1}{n}} > 1,$ then $\sum\limits_{n=1}^{\infty} a_n$ diverges.
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Inconclusive: If $\lim\limits_{n \rightarrow \infty} |a_n|^{\frac{1}{n}} = 1,$ then the test is inconclusive.
As with the root test, the ratio test can be generalized by using the limit supremum to cover cases where the root's limit does not exist but its limit superior or inferior do:
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Convergence: If $\limsup\limits_{n \rightarrow \infty} \sqrt[n]{|a_n|} < 1,$ then $\sum\limits_{n=1}^{\infty} a_n$ converges.
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Divergence: If $\limsup\limits_{n \rightarrow \infty} |a_n|^{\frac{1}{n}} > 1,$ then $\sum\limits_{n=1}^{\infty} a_n$ diverges.
As with all of the other tests, the root test provides no path towards the actual value of a limit, only its existence.
The Alternating Series Test
An alternating series is a series of the form
$$\sum\limits_{n=1}^{\infty} (-1)^na_n$$
where $\{a_n\}$ is a sequence of either all positive or all negative real numbers. The alternating series test states that an alternating sequence converges if $\{|a_n|\}$ is monotonically decreasing and $\lim\limits_{n \rightarrow \infty} a_n = 1.$
The Telescoping Series Test
If $\{a_n\}$ is a sequence, then the series $\sum\limits_{n=1}^{\infty} a_n - a_{n+1}$ is a telescoping series. The name derives from the fact that the term negated in step $n$ of the series is added back in during step $n+1,$ giving the effect of collapsing the interior terms of the sum inwards like a telescope:
$$\sum\limits_{n=1}^{\infty} a_n - a_{n+1} = a_1 - a_2 + a_2 + a_3 - a_3 + \ldots$$
The telescoping series test states that if the sequence $\{a_n\}$ converges, then the series $\sum\limits_{n=1}^{\infty} a_n - a_{n+1}$ converges.
Abel's Test
Abel's test states the that if $\sum\limits_{n=1}^{\infty} a_n$ converges and $\{b_n\}$ is a monotonic and bounded sequence, then $\sum\limits_{n=1}^{\infty} a_nb_n$ also converges.
Problems
Absolute convergence: Show that if the partial sums of a series are absolutely convergent, then the series is convergent.
Let $\sum\limits_{n=1}^{\infty} |a_n|$ be a convergent series. Define $\{s_n\}$ as the sequence of partial sums where $s_n = a_1 + \ldots + a_n,$ and define $\{t_n\}$ as the sequence of partial sums of the absolute values where $t_n = |a_1| + \ldots + |a_n|.$ Because $\{t_n\}$ converges, it is also a Cauchy sequence. Thus, for any $\varepsilon > 0$ there exists an $N \in \mathbb{N}$ such that $|t_m - t_n| < \varepsilon$ whenever $m > n \geq N.$ By the triangle inequality, we see that
$|t_m - t_n| \leq |a_{n+1}| + \ldots + |a_{m}| < \varepsilon$
The triangle inequality also shows that
$|s_m - s_n| = |a_{n+1} + \ldots + a_{m}| \leq |a_{n+1}| + \ldots + |a_{m}| < \varepsilon$
It follows that $\{s_n\}$ is also Cauchy and therefore converges.
Comparison convergence test: Show that if $0 \leq a_n \leq b_n$ for all $n \in \mathbb{N}^+$ and $\sum\limits_{n=1}^{\infty} b_n$ converges, then $\sum\limits_{n = 1}^{\infty} a_n$ converges.
Assume $0 \leq a_n \leq b_n$ for all $n \in \mathbb{N}^+$ and that $\sum\limits_{n=1}^{\infty} b_n$ converges. Let $s_n = \sum\limits_{k=1}^{n} a_k$ and $t_n = \sum\limits_{k=1}^{n} b_k$ be the partial sums. Since $\sum\limits_{n=1}^{\infty} b_n$ converges, there exists an $L$ such that $\lim\limits_{n \rightarrow \infty} t_n = L.$ Since $a_n \leq b_n$ for all $n \in \mathbb{N},$ it follows that $s_n \leq t_n \leq L$ for all $n \in \mathbb{N}^+.$ Since $a_n \geq 0,$ we see that $\{s_n\}$ is monotonically increasing. It follows by the monotone convergence theorem that $s_n$ converges. Therefore $\sum\limits_{n=1}^{\infty} a_n$ converges.
Comparison divergence test: Show that if $0 \leq a_n \leq b_n$ for all $n \in \mathbb{N}$ and $\sum\limits_{n=1}^{\infty} a_n$ diverges, then $\sum\limits_{n=1}^{\infty} b_n$ also diverges.
Assume $0 \leq a_n \leq b_n$ for all $n \in \mathbb{N}$ and that $\sum\limits_{n=1}^{\infty} a_n$ diverges. Let $s_n = \sum\limits_{n=1}^{n} a_n$ and $t_n = \sum\limits_{n=1}^{n} b_n$ be the partial sums. Since $0 \leq a_n \leq b_n$ for all $n \in \mathbb{N},$ it follows that $\{s_n\}$ and $\{t_n\}$ are both monotonically increasing. Because $\sum\limits_{n=1}^{\infty} a_n$ diverges, it follows that $\{s_n\}$ does not converge. By the monotone convergence theorem, we see that $\{s_n\}$ is not bounded. Since $a_n \leq b_n,$ it follows hat $s_n < t_n,$ and therefore $\{t_n\}$ is not bounded. By the monotone convergence theorem again, we conclude that $\{t_n\}$ does not converge. Therefore $\sum\limits_{n=1}^{\infty} b_n$ diverges.
Comparison limit test: Show that if $0 < \lim\limits_{n \rightarrow \infty} \dfrac{a_n}{b_n} < \infty$ and $a_n \geq 0$ and $b_n > 0$ for all $n \in \mathbb{N}^+,$ then either $\sum\limits_{n=1}^{\infty} a_n$ and $\sum\limits_{n=1}^{\infty} b_n$ both converge or diverge.
Assume $a_n \geq 0$ and $b_n > 0$ for all $n \in \mathbb{N}^+,$ and assume that $0 < \lim\limits_{n \rightarrow \infty} \frac{a_n}{b_n} = L.$ Then for any $\varepsilon > 0$ there is some $N \in \mathbb{N}^+$ such that $\left|\frac{a_n}{b_n} - L\right| < \varepsilon.$ Thus $L - \varepsilon < \frac{a_n}{b_n} < L + \varepsilon.$ Multiplying by $b_n$ gives $(L - \varepsilon)b_n < a_n < (L+\varepsilon)b_n.$ By the comparison divergence test, if $\sum\limits_{n=1}^{\infty} b_n$ diverges, then so does $\sum\limits_{n=1}^{\infty} a_n,$ for $(L-\varepsilon)b_n < a_n.$ Likewise, by the comparison convergence test, if $\sum\limits_{n=1}^{\infty} b_n$ converges, then so does $\sum\limits_{n=1}^{\infty} a_n,$ for $a_n < (L+\varepsilon)b_n.$ Assuming instead the divergence or convergence of $\sum\limits_{n=1}^{\infty} a_n$ proves the corresponding divergence or convergence of $\sum\limits_{n=1}^{\infty} b_n$ by the same logic.
Generalized harmonic series: Show that $\sum\limits_{n=1}^{\infty} \dfrac{1}{n^p}$ diverges when $0 < p < 1.$
Note that $\dfrac{1}{n^p} > \dfrac{1}{n}$ when $0 < p < 1.$ Since the harmonic series diverges, it follows by the comparison divergence test that $\sum\limits_{n=1}^{\infty} \dfrac{1}{n^p}$ diverges.
Cauchy's condensation test: Let $\{a_n\}$ be a positive, monotonically decreasing sequence. Show that $\sum\limits_{n=1}^{\infty}a_n$ converges if and only if $\sum\limits_{n=1}^{\infty}2^na_{2^n}$ converges.
Assume $\sum\limits_{n=1}^{\infty}a_n$ converges. Because $\{a_n\}$ is monotonically decreasing, it follows that $a_{2^n} \leq a_n.$
Consider the partial sums $s_n = \sum\limits_{k=1}^{n} a_n$ and $t_n = \sum\limits_{k=1}^{n} 2^ka_{2k}.$ When $n < 2^k,$ we note that, because $a_n$ is monotonically decreasing,
$s_n \leq a_1 + (a_2 + a_3) + \ldots (a_{2^k} + \ldots a_{2^{k+1}-1})\\s^n \leq a_1 + 2^ka_2 + \ldots 2^ka_{2^k}\\s_n \leq t_k$
Therefore $s_n \leq t_k$ when $n < 2^k.$ Likewise, when $n > 2^k,$ we note that
$s_n \geq a_1 + a_2 + (a_3 + a_4) + \ldots (a_{2^{k-1}+1} + \ldots + a_{2^k})\\s_n \geq \dfrac{1}{2}a_1 + a_2 + 2a_4 + \ldots + 2^{k-1}a_{2^k}\\s_n \geq \dfrac{1}{2}t_k$
Therefore $2s_n \geq t_k.$ Together we note that $s_n \leq t_k \leq 2s_n$ when $n < 2^k.$ Therefore if $\{s_n\}$ converges then so does $\{t_k\},$ and likewise if either diverges then so does the other.
Generalized harmonic series: Show that $\sum\limits_{n=1}^{\infty} \dfrac{1}{n^p}$ converges when $p > 1.$
Hint: Use the Cauchy condensation test.
Cauchy's condensation test states that $\sum\limits_{n=1}^{\infty}$ converges if and only if $\sum\limits_{n=1}^{\infty}2^na_{2^n}$ converges. Therefore we investigate whether $\sum\limits_{n=1}^{\infty}2^n \dfrac{1}{2^{np}}$ converges. This series simplifies to $\sum\limits_{n=1}^{\infty} 2^{(1-p)n}.$ Note that $2^{1-p} < 1$ only when $p > 1.$ In this case, we note that $\sum\limits_{n=1}^{\infty} 2^{(1-p)n}$ becomes a convergent geometric series. It follows by the Cauchy condensation test that $\sum\limits_{n=1}^{\infty} \dfrac{1}{n^p}$ converges.
Ratio convergence test: Let $\{a_n\}$ be a sequence of nonzero real numbers. Show that if $\lim\limits_{n \rightarrow \infty} \left|\dfrac{a_{n+1}}{a_n}\right| < 1,$ then $\sum\limits_{n=1}^{\infty} a_n$ converges.
Let $\{a_n\}$ be a sequence of nonzero real numbers, and assume that $\lim\limits_{n \rightarrow \infty} \left|\dfrac{a_{n+1}}{a_n}\right| = L < 1.$ Then for every $\varepsilon > 0$ there exists an $N \in \mathbb{N}^+$ such that $\left|\dfrac{a_{n+1}}{a_n}\right| - L < \varepsilon$ whenever $n \geq N.$ Some algebra shows that $\left|a_{n+1}\right| < (L + \varepsilon)\left|a_n\right|.$ It follows that $\left|a_{N+k}\right| < (L + \varepsilon)^k\left|a_N\right|.$ The absolute inequality for any $n \geq N$ is therefore $\left|a_n\right| < (L+\varepsilon)^{n-N}\left|a_N\right|.$ Since we can select $\varepsilon$ arbitrarily, we may choose it such that $L + \varepsilon < 1.$ Therefore $\lim\limits_{n \rightarrow \infty} (L + \varepsilon)^k = 0.$ The algebraic limit theorem shows that $\lim\limits_{n \rightarrow \infty} |a_N|(L + \varepsilon)^k = 0$ as well. Because $L + \varepsilon < 1,$ we know that the geometric series $\sum\limits_{n=1}^{\infty} (L + \varepsilon)^n$ converges. The algebraic limit theorem for series shows that $\sum\limits_{n=1}^{\infty} |a_N|(L + \varepsilon)^n$ converges as well. From here, the comparison test lets us see that $\sum\limits_{n=1}^{\infty} |a^n|$ converges. Since absolute convergence implies converges, we conclude that $\sum\limits_{n=1}^{\infty} a^n$ converges.
Ratio divergence test: Let $\{a_n\}$ be a sequence of nonzero real numbers. Show that if $\lim\limits_{n \rightarrow \infty} \left| \dfrac{a_{n+1}}{a_n} \right| > 1,$ then $\sum\limits_{n=1}^{\infty} a^n$ diverges.
Let $\{a_n\}$ be a sequence of nonzero real numbers, and assume $\lim\limits_{n \rightarrow \infty} \left|\dfrac{a_{n+1}}{a_n} \right| \geq 1.$ Then there is some $N \in \mathbb{N}^+$ such that $\left|\dfrac{a_{n+1}}{a_n}\right| > 1$ whenever $n \geq N.$ Therefore $|a_{n+1}| > |a_n|.$ It follows that $\lim\limits_{n \rightarrow \infty} a_n \neq 0.$ Therefore $\sum\limits_{n=1}^{\infty} a_n$ diverges.
Counterexample: Provide an example of a series such that $\lim\limits_{n \rightarrow \infty} \left|\dfrac{a_{n+1}}{a_n}\right| = 1$ and $\sum\limits_{n=1}^{\infty} a_n$ converges.
Consider the sequence $a_n = \dfrac{1}{n^2}.$ We seen $\lim\limits_{n \rightarrow \infty} \left|\dfrac{n^2}{(n+1)^2}\right| = \lim\limits_{n \rightarrow \infty} \dfrac{n^2}{n^2+2n+1} = 1.$ As shown above, the generalized harmonic series $\sum\limits_{n=1}^{\infty} \dfrac{1}{n^2}$ converges.
Counterexample: Provide an example of a series such that $\lim\limits_{n \rightarrow \infty} \left|\dfrac{a_{n+1}}{a_n}\right| = 1$ and $\sum\limits_{n=1}^{\infty} a_n$ diverges.
Consider the sequence $a_n = (-1)^n.$ Then $\left|\dfrac{a_{n+1}}{a_n}\right| = \left|\dfrac{(-1)^{n+1}}{(-1)^n}\right| = 1.$ The partial sums $s_n = \sum\limits_{k=1}^{n} (-1)^k = 0$ when $n$ is even and $s_n = -1$ when $n$ is odd. It follows that $\{s_n\}$ diverges, and so $\sum\limits_{n=1}^{\infty} (-1)^n$ diverges.
Generalized ratio convergence test: Show that if $\limsup\limits_{n \rightarrow \infty} \left|\dfrac{a_{n+1}}{a_n}\right| < 1,$ then $\sum\limits_{n=1}^{\infty} a_n$ converges.
Assume $\limsup\limits_{n \rightarrow \infty} \left|\dfrac{a_{n+1}}{a_n}\right| < 1.$ Then there exists for each $r \in \mathbb{R}^+$ where $\limsup\limits_{n \rightarrow \infty} \left|\dfrac{a_{n+1}}{a_n}\right| < r < 1$ an $N \in \mathbb{N}$ such that $\left|\dfrac{a_{n+1}}{a_n}\right| < r$ whenever $n \geq N.$ It follows that $|a_{N+1}| < r|a_N|,$ and in general that $|a_{N+p} |< r^p|a_N|.$ Since $r < 1,$ it follows that $\sum\limits_{n=1}^{\infty} |a_N|r^n = |a_N|\sum\limits_{n=1}^{\infty} r^n$ converges. By the comparison test, it follows that $\sum\limits_{n=1}^{\infty} |a_n|$ converges. By absolute convergence, we conclude that $\sum\limits_{n=1}^{\infty} a_n$ converges.
Generalized ratio divergence test: Show that if there is an $N \in \mathbb{N}$ such that $\left|\dfrac{a_{n+1}}{a_n}\right| \geq 1$ for all $n \geq N,$ then $\sum\limits_{n=1}^{\infty} a_n$ diverges.
Assume there is an $N \in \mathbb{N}$ such that $\left|\dfrac{a_{n+1}}{a_n}\right| \geq 1$ for all $n \geq N.$ Then $|a_{n+1}| \geq |a_n|,$ and so $\lim\limits_{n \rightarrow \infty} a_n \neq 0.$ Therefore $\sum\limits_{n=1}^{\infty} a_n$ diverges.
Generalized ratio divergence test: Show that if $\liminf\limits_{n \rightarrow \infty} \left|\dfrac{a_{n+1}}{a_n}\right| > 1,$ then $\sum\limits_{n=1}^{\infty} a_n$ diverges.
Assume $\liminf\limits_{n \rightarrow \infty} \left|\dfrac{a_{n+1}}{a_n}\right| > 1.$ Then there is some subsequence $\{a_{n_k}\}$ such that $\lim\limits_{k \rightarrow \infty} \left|\dfrac{a_{n_{k+1}}}{a_{n_k}}\right| > 1.$ Therefore there is some $r \in \mathbb{R}$ such that $\lim\limits_{k \rightarrow \infty} \left|\dfrac{a_{n_{k+1}}}{a_{n_k}}\right| > r > 1$ and an $N \in \mathbb{N}$ such that $ \left|\dfrac{a_{n_{k+1}}}{a_{n_k}}\right| > r$ whenever $n \geq N.$ It follows that $|a_{n_{k+1}}| > r|a_{n_k}| > |a_{n_k}|.$ Therefore $\lim\limits_{n \rightarrow \infty} a_{n_k} \neq 0,$ therefore $\lim\limits_{n \rightarrow \infty} a_n \neq 0,$ and so $\sum\limits_{n=1}^{\infty} a_n$ diverges.
Root convergence test: Show that if $\lim\limits_{n \rightarrow \infty} \sqrt[n]{|a_n|} < 1,$ then $\sum\limits_{n=1}^{\infty} a_n$ converges.
Assume $\lim\limits_{n \rightarrow \infty} \sqrt[n]{|a_n|} = a < 1.$ Then for every $\varepsilon > 0$ there is some $N \in \mathbb{N}$ such that $|\sqrt[n]{|a_n|} - 1| < \varepsilon$ whenever $n \geq N.$ Pick $\varepsilon$ such that $a < a + \varepsilon < 1,$ and set $b = a + \varepsilon$ for convenience. It follows that $|a_n| < b^n$ whenever $n \geq N.$ Since $0 < b < 1,$ the geometric series $\sum\limits_{n=1}^{\infty} b^n$ converges. By the comparison test, it follows that $\sum\limits_{n=1}^{\infty} |a_n|$ converges, and by absolute convergence, $\sum\limits_{n=1}^{\infty} a_n$ therefore also converges.
Root divergence test: Show that if $\lim\limits_{n \rightarrow \infty} \sqrt[n]{|a_n|} > 1,$ then $\sum\limits_{n=0}^{\infty} a_n$ diverges.
Assume $\lim\limits_{n \rightarrow \infty} \sqrt[n]{|a_n|} = a > 1.$ Then for every $\varepsilon > 0$ there exists an $N \in \mathbb{N}$ such that $|\sqrt[n]{|a_n|} - a| < \varepsilon$ whenever $n \geq N.$ Let $b = a - \varepsilon$ and pick $\varepsilon$ such that $a - \varepsilon > 1.$ Then $1 < b < \sqrt[n]{|a_n|}$ whenever $n \geq N.$ It follows that $1 < b^n < |a_n|.$ Since $|a_n| > 1$ for infinitely many values of $n,$ it follows that $\lim\limits_{n \rightarrow \infty} |a_n| \neq 0,$ and therefore $\sum\limits_{n=0}^{\infty} a_n$ diverges.
Counterexample: Provides an example of a sequence $\{a_n\}$ such that $\lim\limits_{n \rightarrow \infty} \sqrt[n]{|a_n|} = 1$ and $\sum\limits_{n=1}^{\infty} a_n$ diverges.
Consider the series $\sum\limits_{n=1}^{\infty} 1.$ Then $\lim\limits_{n \rightarrow \infty} \sqrt[n]{\left|1\right|} = 1,$ and $\sum\limits_{n=1}^{\infty} 1$ diverges.
Generalized root convergence test: Show that if $\limsup\limits_{n \rightarrow \infty} \sqrt[n]{|a_n|} < 1,$ then $\sum\limits_{n \rightarrow \infty} a_n$ converges.
Assume $\limsup\limits_{n \rightarrow \infty} \sqrt[n]{|a_n|} = a < 1.$ Pick $b$ such that $a < b < 1.$ By the convergence of the limit superior, there exists an $N \in \mathbb{N}$ such that $\sqrt[n]{|a_n|} < b$ for all $n \geq N.$ Therefore $|a_n| < b^n.$ Since $0 < b < 1,$ the geometric series $\sum\limits_{n=0}^{\infty} b^n$ converges. It follows by the comparison test that $\sum\limits_{n=0}^{\infty} |a_n|$ also converges, and by absolute convergence that $\sum\limits_{n=0}^{\infty} a_n$ converges.
Generalized root divergence test: Show that if $\limsup\limits_{n \rightarrow \infty} \sqrt[n]{a_n} > 1,$ then $\sum\limits_{n=0}^{\infty} a_n$ diverges.
Assume $\limsup\limits_{n \rightarrow \infty} \sqrt[n]{a_n} = p > 1.$ Then there exists some subsequence $\{a_{n_k}\}$ such that $\lim\limits_{k \rightarrow \infty} \sqrt[n_k]{a_{n_k}} = p.$ Therefore $a_n$ does not converge to $0,$ and so $\sum\limits_{n=0}^{\infty} a_n$ does not converge.
Alternating series test: Show that if $\{a_n\}$ is an alternating series, $\{|a_n|\}$ is monotonically decreasing, and $\lim\limits_{n \rightarrow \infty} a_n = 0,$ then $\sum\limits_{n=1}^{\infty} a_n$ converges.
Hint: Consider a finite partition of $\{a_n\}.$
First, assume all the terms of $\{a_n\}$ are positive - the proof of the alternate case is identical.
Consider the partial sums of an even number of terms, $s_{2n} = \sum\limits_{k=1}^{2n} a_k,$ and an odd number of terms, $s_{2n+1} = \sum\limits_{k=1}^{2n+1} a_n.$
The odd numbered partial sums decrease monotonically, as $s_{2(n+1)+1} = s_{2n+1} - a_{2n+2} + a_{2n+3} \leq s_{2n+1}.$ Likewise, the even numbered partial sums increase monotonically, as $s_{2(n+1)} = s_{2n} + a_{2n+1} - a_{2n+2} \leq s_{2n}.$ Because all the terms of $\{a_n\}$ are positive, it follows that $s_{2n+1} - s_{2n} = a_{2n+1} \geq 0.$
Next, note that $a_1 - a_2 = s_2 \leq s_{2n} \leq s_{2n+1} \leq s_1 = a_1.$ Because $a_1 - a_2$ is a lower bound for $\{s_{2n+1}\},$ it follows by the monotone convergence theorem that $\{s_{2n+1}\}$ converges. Likewise, because $a_1$ is an upper bound for $\{s_{2n}\},$ it follows by the monotone convergence theorem that $\{s_{2n}\}$ also converges. Finally, note that $s_{2n+1} - s_{2n} = a_{2n+1}.$ Since every subsequence of a convergent sequence converges to the same limit, we can see that $\lim\limits_{n \rightarrow \infty} a_{2n+1} = 0.$ It therefore follows that $\lim\limits_{n \rightarrow \infty} s_{2n+1} - s_{2n} = 0.$ We conclude by the algebraic limit theorem that $\lim\limits_{n \rightarrow \infty} s_{2n} = \lim\limits_{n \rightarrow \infty} s_{2n+1}.$ Because $\{s_{2n}\}$ and $\{s_{2n+1}\}$ form a finite partition of subsequences of $\{s_m\},$ it follows that $\{s_m\}$ converges, and therefore $\sum\limits_{n=1}^{\infty} a_n$ converges.
Telescoping series test: Show that if $\{a_n\}$ is monotonic and bounded, then $\sum\limits_{n=1}^{\infty} a_n - a_{n+1} = a_1 - \lim\limits_{n \rightarrow \infty} a_n.$
Let $\{s_n\}$ be the sequence of partial sums of $a_n.$ Then $s_n = a_1 - a_{n+1}.$ Since $\{a_n\}$ is monotonic and bounded, it converges by the monotone convergence theorem. Therefore $\lim\limits_{n \rightarrow \infty} s_n = \lim\limits_{n \rightarrow \infty} a_1 - a_{n+1} = a_1 - \lim\limits_{n \rightarrow \infty} a_n.$
Abel's test: Show that if $\sum\limits_{n=1}^{\infty} a_n = L$ and $\{b_n\}$ is monotonic and bounded, then $\sum\limits_{n=1}^{\infty} a_nb_n$ converges.
Hint: Use summation by parts.
Let $\{s_n\}$ be the sequence of partial sums of $\{a_n\},$ and let $\sum\limits_{n=1}^{\infty} a_n = L.$ Because $\{b_n\}$ is monotonic and bounded, it follows by the monotone convergence theorem that $\{b\}$ converges. Let $\lim\limits_{n \rightarrow \infty} b_n = M.$ Summation by parts shows the following:
$ \sum\limits_{n=1}^{m} a_mb_m = s_mb_m - s_0b_1 + \sum\limits_{n=1}^{m-1}s_n(b_n - b_{n+1}) \\ \sum\limits_{n=1}^{m} a_mb_m = s_mb_m - 0 \cdot b_1 + \sum\limits_{n=1}^{m-1}s_n(b_n - b_{n+1}) \\ \sum\limits_{n=1}^{m} a_mb_m = s_mb_m + \sum\limits_{n=1}^{m-1}s_n(b_n - b_{n+1}) \\ \lim\limits_{m \rightarrow \infty} \sum\limits_{n=1}^{m} a_mb_m = \lim\limits_{m \rightarrow \infty} \left(s_mb_m + \sum\limits_{n=1}^{m-1}s_n(b_n - b_{n+1}) \right) \\ \sum\limits_{n=1}^{\infty} a_mb_m = \lim\limits_{m \rightarrow \infty} (s_mb_m) + \lim\limits_{m \rightarrow \infty} \sum\limits_{n=1}^{m-1}s_n(b_n - b_{n+1}) \\ \sum\limits_{n=1}^{\infty} a_mb_m = (\lim\limits_{m \rightarrow \infty}s_m)(\lim\limits_{m \rightarrow \infty}b_m) + \sum\limits_{n=1}^{\infty}s_n(b_n - b_{n+1}) \\ \sum\limits_{n=1}^{\infty} a_mb_m = LM + \sum\limits_{n=1}^{\infty}s_n(b_n - b_{n+1}) \\ $
We consider the absolute convergence of the summation on the righthand side of the equation:
$ \sum\limits_{n=1}^{\infty}|s_n(b_n - b_{n+1})| = \sum\limits_{n=1}^{\infty}|s_n||b_n-b_{n+1}| \\ \sum\limits_{n=1}^{\infty}|s_n(b_n - b_{n+1})| \leq \sum\limits_{n=1}^{\infty}|L||b_n-b_{n+1}| \\ \sum\limits_{n=1}^{\infty}|s_n(b_n - b_{n+1})| \leq |L|\sum\limits_{n=1}^{\infty}|b_n-b_{n+1}| $
If $\{b_n\}$ is monotonically decreasing, the series is a telescoping series:
$ |L|\sum\limits_{n=1}^{\infty}|b_n-b_{n+1}| = |L|\sum\limits_{n=1}^{\infty}b_n-b_{n+1} \\ |L|\sum\limits_{n=1}^{\infty}|b_n-b_{n+1}| = |L|(b_1 - M) $
If $\{b_n\}$ is monotonically increasing, the series is also a telescoping series:
$ |L|\sum\limits_{n=1}^{\infty}|b_n-b_{n+1}| = |L|\sum\limits_{n=1}^{\infty}-(b_n-b_{n+1}) \\ |L|\sum\limits_{n=1}^{\infty}|b_n-b_{n+1}| = |L|(M-b_1) $
In either case it follows by the comparison test that $\sum\limits_{n=1}^{\infty}|s_n(b_n - b_{n+1})|$ converges. It then follows by absolute convergence that $\sum\limits_{n=1}^{\infty}s_n(b_n - b_{n+1}).$ Therefore $\sum\limits_{n=1}^{\infty} a_nb_n$ converges.