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I need to prove that:

$$\sum_{n=1}^{\infty} (-1)^{n+1}\log\left(1+\frac{1}{n}\right)$$

is convergent, but not absolutely convergent.

I tried the ratio test:

$$\frac{a_{n+1}}{a_n} = -\frac{\log\left(1+\frac{1}{n+1}\right)}{\log\left(1+\frac{1}{n}\right)} = -\log\left({\frac{1}{n+1}-\frac{1}{n}}\right)$$

I know that the thing inside the $\log$ converges to $1$, so $-\log$ converges to $0$? This is not right, I cannot conclude that this series is divergent.

Also, for the sequence without the $(-1)^{n+1}$ it would give $0$ too.

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First, because is an alternating series, we start with the alternating series test: if $(a_n)\to 0$ and $|a_n|$ decreases then $\sum(-1)^{n}a_n$ converges.

We can see that $\log\left(\frac{n+1}{n}\right)\to 0$ because $\left(\frac{n+1}{n}\right)\to 1$ and $\left|\log\left(\frac{n+1}{n}\right)\right|$ decreases because logarithm is a strict increasing function and $\frac{n+1}{n}$ obviously decreases. Then the series, at least, is conditionally convergent.

To test absolute convergence we can see that we can write

$$\sum_{n=1}^{\infty}\log\left(\frac{n+1}{n}\right)=\sum_{n=1}^{\infty}\log(n+1)-\log(n)$$

that is a telescoping sum. And because $\sum_{n=1}^{\infty}a_n=\lim_{N\to\infty}\sum_{n=1}^{N}a_n$ then

$$\sum_{n=1}^{\infty}\log(n+1)-\log(n)=\lim_{N\to\infty}\sum_{n=1}^{N}\log(n+1)-\log(n)=\lim_{N\to\infty}\log(N+1)$$

what is divergent, so the series is conditionally convergent.

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$$(\text{Telescopic sum})\quad\sum_{n=1}^{N}\log\frac{n+1}{n} = \log(N+1)\tag{1}$$

$$\sum_{n=1}^{2N}(-1)^{n+1}\frac{\log(n+1)}{\log n} = \log\prod_{n=1}^{N}\frac{4n^2}{4n^2-1}=-\log\prod_{n=1}^{N}\left(1-\frac{1}{4n^2}\right)\tag{2} $$

and the last product is convergent since $\sum_{n\geq 1}\frac{1}{4n^2}$ is convergent. In facts, the limit of the RHS of $(2)$ as $N\to +\infty$ is just the logarithm of Wallis' product.

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Hint: for the convergence, use the alternating test

https://en.wikipedia.org/wiki/Alternating_series_test

It does not converge absolutely since $log(1+1/n)\simeq 1/n$

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First i would show the absolute converge by the test for alternating series. First show that it is a alternating series (pretty obvious since your log only gives positive values). Then since $$ log(1+ \frac{1}{n} ) \rightarrow 0 \ , n \rightarrow \infty $$ it is convergent. To show the absolute i would remove $ (-1)^{n+1} $ since its only changing +-. Then i would use comparison test at compare to the divergent series $$ \frac{1}{n} $$

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  • $\begingroup$ How you compare with $1/n$? If you compare both we can see that $1/n>\ln(n+1)-\ln n$, and then what. $\endgroup$ – Masacroso May 29 '16 at 0:45

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