The $n$th-Term Test for a Divergent Series
Theorem 1. If $\sum_{n=1}^{\infty} a_n$ converges, then $a_n \rightarrow 0$.
Proof. Let $\sum_{n=1}^{\infty} a_n = \lim_{n \to \infty} s_n = L$, where $s_n$ is the partial sums of the series and $L$ is the sum of the series. Note that $$\lim_{n \to \infty} s_n = \lim_{n \to \infty} s_{n-1} = L \\ \Longrightarrow \lim_{n \to \infty} (s_n - s_{n-1}) = \lim_{n \to \infty} a_n = L - L = 0. \blacksquare$$
Corollary. $\sum_{n=1}^{\infty} a_n$ diverges if $\lim_{n \to \infty} a_n$ fails to exists or is different from zero.
어떤 수열의 series의 수렴성을 판정하기 위해 유용하게 쓰이는 test 중 하나로, series가 수렴한다면 반드시 그 수열은 $n$이 아주 큰 상황에서는 0으로 줄어드는 거동을 보여야 한다는 정리다.
The Integral Test
Theorem 2. Let $\{ a_n\}$ be a sequence of positive terms. Suppose that $a_n = f(n)$, where $f$ is a continuous, positive, decreasing function of $x$ for all $x \geq N$ ($N$ a positive integer). Then the series $\sum_{n = N}^{\infty} a_n$ an and the integral $\int_N^{\infty} f(x) dx$ both converge or both diverge.
Proof. It is true that $$\int_N^{n+1} f(x) dx \leq a_N + a_{N+1} + \cdots + a_{n}$$ for every $n \geq N$ if we regard each $a_n$ the rectangles with the height $a_n$ and the width 1. Similarly, we can also see that $$a_{N+1} + a_{N+2} + \cdots + a_n \leq \int_N^n f(x) dx.$$ Combining these results gives $$\int_N^{n+1} \leq a_N + \cdots + a_n \leq a_N + \int_N^n f(x) dx.$$ These inequalities hold for each $n \leq N$, and continue to hold as $n \to \infty$. Therefore, if $\int_{N}^{\infty} f(x) dx$ converges, then $\sum_{n=N}^{\infty} a_n$ also converges. Conversely, if $\int_N^{\infty} f(x) dx$ diverges, then $\sum_{n=N}^{\infty} a_n$ also diverges. $\blacksquare$
The $p$-Series Test
Corollary of Theorem 2. The $p$-series $$\sum_{n=1}^{\infty} \frac{1}{n^p}$$ converges if $p > 1$, and diverges if $p \leq 1$.
$p$-series test로부터 조화급수 $\sum_{n=1}^{\infty} \frac{1}{n}$이 발산함을 알 수 있다. 재밌는 것은 조화급수는 $p=1$일 때의 경우인데, 1은 정확히 $p$-series test의 기준점이다. 실제로 계산해보면 조화급수는 매우, 매우, 매우 느리게 발산하는 것을 알 수 있다. (178만 번째 항까지 더해야 겨우 20을 넘긴다)
The Comparison Test
Theorem 3. Let $\sum a_n$ and $\sum b_n$ be two series with $0 \leq a_n \leq b_n$ for all $n$. Then
(1) If $\sum b_n$ converges, then $\sum a_n$ also converges.
(2) If $\sum a_n$ diverges, then $\sum b_n$ also diverges.
Proof. Let $\sum b_n = L$. Then the partial sums of $a_n$ is bounded above, since $0 \leq \sum_{n=1}^{k} a_n \leq \sum_{n=1}^k b_n \leq L$. Then $\sum a_n$ converges by Theorem 2. (2) is proved from the contrapositive of (1). $\blacksquare$
이미 결과를 알거나 다른 방법을 통해 결과를 알 수 있는, 더 간단한 형태의 급수를 이용해 수렴 여부를 계산할 수 있는 방법이다.
The Limit Comparison Test
Theorem 4. Suppose that $a_n > 0$ and $b_n > 0$ for all $n \geq N$.
(1) If $\lim_{n \to \infty} \frac{a_n}{b_n} = c > 0$, then $\sum a_n$ and $\sum b_n$ both converge or both diverge.
(2) If $\lim_{n \to \infty} \frac{a_n}{b_n} = 0$ and $\sum b_n$ converges, then $\sum a_n$ converges.
(3) If $\lim_{n \to \infty} \frac{a_n}{b_n} = \infty$ and $\sum b_n$ diverges, then $\sum a_n$ diverges.
Proof. (1) Since $\frac{c}{2} > 0$, there exists $N$ such that for $n \geq N$, $|\frac{a_n}{b_n} - c| < \frac{c}{2}$. Then for $n \geq N$, we have $$\frac{c}{2} < \frac{a_n}{b_n} < \frac{3c}{2} \\ \Longrightarrow \frac{c}{2} \cdot b_n < a_n < \frac{3c}{2} \cdot b_n.$$ If $\sum b_n$ converges, then $\sum \frac{3c}{2} b_n$ converges and $\sum a_n$ also converges by the Comparison Test. If $\sum b_n$ diverges, then $\sum \frac{c}{2} b_n$ diverges and $\sum a_n$ also diverges by the Comparison Test.
(2) and (3) can be proved similar to (1). $\blacksquare$
Absolute Convergence
Definition 1. A series $\sum a_n$ is absolutely convergent (converges absolutely) if the corresponding series of absolute values, $\sum |a_n|$, converges.
Conditional Convergence
Definition 2. A series that is convergent but not absolutely convergent is called conditionally convergent (converges conditionally).
The Absolute Convergence Test
Theorem 5. If $\sum_{n=1}^{\infty} |a_n|$ convergent, then $\sum_{n=1}^{\infty} a_n$ converges.
Proof. Note that $a_n \leq |a_n|, \forall n \in \mathbb{N}$. Since $\sum_{n=1}^{\infty} |a_n|$ convergent, by the Comparison Test, we have $\sum_{n=1}^{\infty} a_n$ converges. $\blacksquare$
The Ratio Test
Theorem 6. Let $\sum a_n$ be any series and suppose that $$\lim_{n \to \infty} \bigg| \frac{a_{n+1}}{a_n} \bigg| = \rho.$$ Then
(1) the series is absolutely convergent if $\rho < 1$,
(2) the series diverges if $\rho > 1$ or $\rho$ is infinite.
(3) the test is inconclusive if $\rho = 1$.
Proof. (1) Let's take $r$ by $\rho < r < 1$. Then the number $\varepsilon = r - \rho$ is positive. Since $$\bigg| \frac{a_{n+1}}{a_n} \bigg| \to \rho,$$ we have $$\bigg| \frac{a_{n+1}}{a_n} \bigg| < \rho + \varepsilon = r$$ for every $n \geq N$ where $N$ is some natural number. Then $$|a_{N+1}| < r|a_N| \\ \Longrightarrow |a_{N+2}| < r|a_{N+1}| < r^2|a_N| \\ \Longrightarrow |a_{N+3}| < r|a_{N+2}| < r^3|a_{N}| \\ \vdots \\ |a_{N+m}| < r|a_{N+m-1}| < r^m|a_N|.$$ Therefore, $$\sum_{m=N}^{\infty} |a_m| = \sum_{m=0}^{\infty} |a_{N+m}| \leq \sum_{m=0}^{\infty} r^m |a_N| = |a_N| \sum_{m=0}^{\infty} r^m.$$ Because $0 < r < 1$, the right-hand side of the inequality is convergent. This means, by the Comparison Theorem, that $\sum_{m=N}^{\infty} |a_m|$ is convergent. Since finitely many terms do not affect the convergence of the series, the series $\sum_{m=1}^{\infty} a_m$ is absolutely convergent.
(2) If $\rho > 1$, then we take $r$ by $\rho > r > 1$ and set $\varepsilon = \rho - r > 0$. Then for $n \geq N$ where $N$ is some natural number, $$\bigg| \bigg| \frac{a_{n+1}}{a_n} \bigg| - \rho \bigg| = \bigg| \rho - \bigg| \frac{a_{n+1}}{a_n}\bigg| \bigg| < \varepsilon \\ \Longrightarrow 1 < r = \rho - \varepsilon < \bigg| \frac{a_{n+1}}{a_n} \bigg|.$$ Similarly, we obtain $$1 < \bigg| \frac{a_{n+1}}{a_n} \bigg|$$ for every $n \leq N$ if $\rho = \infty$. Therefore, we have $$|a_N| < |a_{N+1}| < |a_{N+2}| < \cdots .$$ This means that $\sum a_n$ does not converges absolutetly.
(3) Compare two series, $\sum_{n=1}^{\infty} \frac{1}{n}$ and $\sum_{n=1}^{\infty} \frac{1}{n^2}$. $\blacksquare$
The Root Test
Theorem 7. Let $\sum a_n$ be any series and suppose that $$\lim_{n \to \infty} \sqrt[n]{ |a_n| } = \rho.$$ Then
(1) the series converges absolutely if $\rho < 1$,
(2) the series diverges if $\rho > 1$ or $\rho$ is infinite,
(3) the test is inconclusive if $\rho = 1$.
Proof. (1) Take $\varepsilon > 0$ by $\rho + \varepsilon < 1$. Then there exists $M \in \mathbb{N}$ such that for all $n \geq M$, $$|\sqrt[n]{|a_n|} - \rho| < \varepsilon \\ \Longrightarrow \sqrt[n]{|a_n|} < \rho + \varepsilon \\ \Longrightarrow |a_n| < (\rho + \varepsilon)^n.$$ Since $\sum_{n=N}^{\infty} (\rho + \varepsilon)^n$ is a geometric series and $(\rho + \varepsilon) < 1$, the series converges. Then, by the Comparison Theorem, $\sum_{n=N}^{\infty} |a_n|$ converges and $\sum_{n=1}^{\infty} |a_n|$ also converges.
(2) Similar to the proof of the Ratio Test (2), we can say that for every $n \geq N$, $$\sqrt[n]{|a_n|} > 1 \Longrightarrow |a_n| > 1.$$ This means that $\{a_n\}$ does not converge to zero and by the $n$-th Term Test, the series $\sum a_n$ diverges.
(3) Compare two series, $\sum_{n=1}^{\infty} \frac{1}{n}$ and $\sum_{n=1}^{\infty} \frac{1}{n^2}$. $\blacksquare$
The Alternating Series Test
Theorem 8. The series $$\sum_{n=1}^{\infty} (-1)^{n+1} a_n = a_1 - a_2 + a_3 - \cdots$$ converges if the following conditions are satisfied:
(1) $a_n > 0, \forall n \in \mathbb{N}$.
(2) $a_n \geq a_{n+1}$ for all $n \geq N$, for some integer $N$.
(3) $a_n \to 0$.
Proof. Take $N=1$. If $n = 2m$, then the partial sums of the first $n$ terms is $$s_{2m} = (a_1 - a_2) + (a_3 - a_4) + \cdots + (a_{2m-1} - a_{2m}) \\ = a_1 - (a_2 - a_3) - (a_4 - a_5) - \cdots - (a_{2m-2} - a_{2m-1}) - a_{2m}.$$ This means that $s_{2m}$ is monotonically increasing and $s_{2m} \leq a_1$, that is, $s_2m$ is bounded above. Therefore, by the Monotonic Sequence Theorem, $s_{2m}$ converges to some number $L$.
If $n = 2m+1$, then $s_{2m+1} = s_{2m} + a_{2m+1}$. Since $a_n \to 0$, $a_{2m+1} \to 0$. Then we know that $s_{2m+1}$ also converges to $L$. Thus $\lim_{n \to \infty} s_n = \sum_{n=1}^{\infty} a_n = L$. $\blacksquare$
가끔 (2)번 조건, 즉 감소한다는 조건을 보이는 게 어려운 경우가 있는데 이때는 $f(n) = a_n$인 미분가능한 함수 $f$를 정의해 $f'(x) \leq 0$을 보이는 것으로 대체하는 방법도 있다.