Prove Gauss’ test for convergence of a series of positive terms

by RoRi

March 10, 2016

Prove Guass’ test for the convergence of a series $\sum a_n$ with $a_n > 0$ for all $n$ . This test says that if there is an $N \geq 1$ , an $s > 1$ , and an $M > 0$ such that

$\frac{a_{n+1}}{a_n} = 1 - \frac{A}{n} + \frac{f(n)}{n^s} \qquad \text{for} \qquad n \geq N,$

where $|f(n)| \leq M$ for all $n$ , then the series $\sum a_n$ converges if $A > 1$ and diverges if $A \leq 1$ .

Proof. Let $\sum a_n$ be a series of positive terms, and assume there exists an $N \geq 1$ , an $s > 1$ , and an $M > 0$ such that

$\frac{a_{n+1}}{a_n} = 1 - \frac{A}{n} + \frac{f(n)}{n^s}$

for all $n \geq N$ , and $|f(n)| \leq M$ for all $n$ . Now we consider the three cases $A < 1$ , $A > 1$ and $A = 1$ .

Case 1. Assume $A < 1$ . Then we have

$\begin{align*} \frac{a_{n+1}}{a_n} &= 1 - \frac{A}{n} + \frac{f(n)}{n^s} \\[9pt] &= 1 - \frac{1+A-1}{n} + \frac{f(n)}{n^2} \\[9pt] &= 1 - \frac{1}{n} + \frac{1-A}{n} + \frac{f(n)}{n^s}. \end{align*}$

Since $|f(n)| \leq M$ for all $n \geq N$ we have

$\lim_{n \to \infty} \frac{f(n)}{n^s} \leq \lim_{n \to \infty} \frac{M}{n^s} = 0.$

But then since $A < 1$ we have $\frac{1-A}{n} > 0$ and so for all sufficiently large $n$ we have

$\frac{1-A}{n} + \frac{f(n)}{n^s} > 0$

(since we can make $\left| \frac{f(n)}{n^s} \right|$ arbitrarily small so even if $\frac{f(n)}{n^s} < 0$ it is smaller in absolute value than $\frac{1-A}{n}$ and so $\frac{1-A}{n} + \frac{f(n)}{n^s}$ must be positive). Since this term is positive we then have

$\frac{a_{n+1}}{a_n} > 1 - \frac{1}{n}$

for all sufficiently large $n$ . By Raabe’s test (Section 10.16, Exercise #16) we then have $\sum a_n$ diverges.

Case 2. Assume $A > 1$ . Then we have

$\begin{align*} \frac{a_{n+1}}{a_n} &= 1 - \frac{A}{n} + \frac{f(n)}{n^s} \\[9pt] &\leq 1 - \frac{A}{n} + \frac{M}{n^s} \\[9pt] &\leq 1 - \frac{1}{n} + \frac{A-1}{n} + \frac{M}{n^s} \\[9pt] &\leq 1 - \frac{1}{n} + \frac{A-1+M}{n} &(\frac{M}{n^s} < \frac{M}{n} \text{ since } s>1)\\[9pt] &\leq 1 - \frac{1}{n} + \frac{r}{n} \end{align*}$

for $r = A - 1 + M > 0$ (since $M>0$ and $A>1$ ). Hence, by Raabe’s test again we have $\sum a_n$ converges.

Case 3. Assume $A = 1$ . From a previous exercise (Section 10.16, Exercise #15) we know that if $\{ a_n \}$ is a sequence of positive terms, $\{ b_n \}$ is a sequence of positive terms such that $\sum \frac{1}{b_n}$ diverges and $c_n \leq 0$ where

$c_n = b_n - \frac{b_{n+1} a_{n+1}}{a_n}$

then $\sum a_n$ diverges. So, in this case, let $b_{n+1} = n \log n$ . Then we know from Example 2 on page 398 of Apostol that $\sum \frac{1}{b_n}$ diverges. So, we let

$\begin{align*} c_n &= b_n - \frac{b_{n+1} a_{n+1}}{a_n} \\[9pt] &= (n-1)\log(n-1) - n \log n \left( 1 - \frac{1}{n} + \frac{f(n)}{n^s} \right) \\[9pt] &= (n-1) \log (n-1) - n \log n \left( \frac{n-1}{n} + \frac{f(n)}{n^s} \right) \\[9pt] &= (n-1) \log (n-1) - \frac{n(n-1)}{n} \log n - \frac{f(n) \log n}{n^{s-1}} \\[9pt] &= (n-1)(\log (n-1) - \log n) - \frac{f(n) \log n}{n^{s-1}}. \end{align*}$

But then, since the logarithm is an increasing function, we know $(n-1)(\log (n-1) - \log n) < 0$ . Furthermore,

$\lim_{n \to \infty} \frac{f(n) \log n}{n^{s-1}} = 0$

since $|f(n)| \leq M$ for all sufficiently large $n$ , and $\lim_{n \to \infty} \frac{\log n}{n^{s-1}} = 0$ by equation (10.11) on page 380 of Apostol. Therefore, for sufficiently large $n$ we have

$(n-1)(\log (n-1) - \log n) - \frac{f(n) \log n}{n^{s-1}} < 0.$

Hence, by the previous exercise, we have $\sum a_n$ diverges.

So, putting all three cases together we have the requested result, $\sum a_n$ converges if $A < 1$ and diverges if $A \geq 1. \qquad \blacksquare$

4 comments

October 15, 2022 at 6:58 am

S says:

Can we simply define a new sequence

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, and let [latex]A = r+1[\latex], to make problems 16 and 17 equivalent (together with application of Teorems 10.2 and 10.3)? To me that solution seems much simpler, especially because there is no special case for [latex]A=1[\latex].

August 14, 2022 at 12:03 pm

Mohammad Azad says:

in the first case you can prove your claim by showing the limit (1-A)/n+f(n)/n^s-1/n=0 and using the definition of limits obtain an inequality and complete the proof.
In the second case, you made an awful mistake that somehow went past you. You can solve it similarly to the first case but by proving that the limit
(1-A)n^s+f(n)=infinity which means it’s positive for any sufficiently large n and complete the proof.
In the third case you can show that the limit of c_n is -1 thus it must be negative for sufficiently large n.

Reply
January 3, 2017 at 1:55 pm

Matt says:

Regarding case $A = 1$ , showing that $\lim_{n \to \infty} \frac{-f(n)\log(n)}{n^{s-1}} = 0$ and $(n-1)\left( \log(n - 1) - \log(n)\right) < 0$ is not sufficient for

$\frac{-f(n)\log(n)}{n^{s-1}} + (n-1)\left( \log(n - 1) - \log(n) \right) < 0$

In other words showing that $a_n < 0$ and $\lim_{n \to \infty} b_n = 0$ does not mean that

$a_n + b_n 0 <img src="https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-42e698812184658dafdec5db754e95e1_l3.png" height="14" width="321" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[Hence, before concluding, we must also note that\]" title="Rendered by QuickLaTeX.com"/> \lim_{n \to \infty} (n-1)\left( \log(n - 1) - \log(n)\right) = -1 <img src="https://www.stumblingrobot.com/wp-content/ql-cache/quicklatex.com-a9aec5f8b810f6025f4f7233ea1c01f8_l3.png" height="43" width="319" class="ql-img-displayed-equation quicklatex-auto-format" alt="\[Which means, for $n > N_1$,\]" title="Rendered by QuickLaTeX.com"/> -\frac{3}{2} < (n-1)\left( \log(n - 1) - \log(n)\right) N_2$

,

$-\frac{1}{2} < \frac{-f(n)\log(n)}{n^{s-1}} < \frac{1}{2}$

So, letting $N = \max\left\{ N_1, N_2 \right\}$

$\frac{-f(n)\log(n)}{n^{s-1}} + (n-1)\left( \log(n - 1) - \log(n)\right)< 0$

which completes the proof.

Reply
- January 3, 2017 at 2:08 pm
 
 Matt says:
 
 I don’t know why it is displayed like that. The Latex is bellow:
 
 Regarding case $A = 1$, showing that $\lim_{n \to \infty} \frac{-f(n)\log(n)}{n^{s-1}} = 0$ and $(n-1)\left( \log(n – 1) – \log(n)\right) < 0$ is not sufficient for
 $$
 \frac{-f(n)\log(n)}{n^{s-1}} + (n-1)\left( \log(n – 1) – \log(n) \right) < 0
 $$
 In other words showing that $a_n < 0$ and $\lim_{n \to \infty} b_n = 0$ does not mean that $a_n + b_n 0
 $$
 Hence, before concluding, we must also note that
 $$
 \lim_{n \to \infty} (n-1)\left( \log(n – 1) – \log(n)\right) = -1
 $$
 Which means, for $n > N_1$,
 $$
 -\frac{3}{2} < (n-1)\left( \log(n – 1) – \log(n)\right) N_2$,
 $$
 -\frac{1}{2} < \frac{-f(n)\log(n)}{n^{s-1}} < \frac{1}{2}
 $$
 So, letting $N = \max\left\{ N_1, N_2 \right\}$
 $$
 \frac{-f(n)\log(n)}{n^{s-1}} + (n-1)\left( \log(n – 1) – \log(n)\right)< 0
 $$
 which completes the proof.
 
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Stumbling Robot

Prove Gauss’ test for convergence of a series of positive terms

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