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

Prove Guass’ test for the convergence of a series with for all . This test says that if there is an , an , and an such that

where for all , then the series converges if and diverges if .

Proof. Let be a series of positive terms, and assume there exists an , an , and an such that

for all , and for all . Now we consider the three cases , and .

Case 1. Assume . Then we have

Since for all we have

But then since we have and so for all sufficiently large we have

(since we can make arbitrarily small so even if it is smaller in absolute value than and so must be positive). Since this term is positive we then have

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

Case 2. Assume . Then we have

for (since and ). Hence, by Raabe’s test again we have converges.

Case 3. Assume . From a previous exercise (Section 10.16, Exercise #15) we know that if is a sequence of positive terms, is a sequence of positive terms such that diverges and where

then diverges. So, in this case, let . Then we know from Example 2 on page 398 of Apostol that diverges. So, we let

But then, since the logarithm is an increasing function, we know . Furthermore,

since for all sufficiently large , and by equation (10.11) on page 380 of Apostol. Therefore, for sufficiently large we have

Hence, by the previous exercise, we have diverges.

So, putting all three cases together we have the requested result, converges if and diverges if

1. 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].

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.

3. Matt says:

Regarding case , showing that and is not sufficient for

In other words showing that and does not mean that

,

So, letting

which completes the proof.

• 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.