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Evaluate: $$\sum_{k=1}^\infty \log(1+\frac{1}{k})$$

So I know that the first one and the last one are stayed because it is telescopically deleted. So,

$$\log\left(1+\frac{1}{k}\right) = \log\left(\frac{k+1}{k}\right) = \log(k+1) - \log(k)$$

And we stay with with $\log(1)$ and $\log(n+1)$ , assuming we got from $k=1$ till $k=n$ to compute the subsequent sums, because I don't know how to evaluate till infinity.

Well $\log(1) = 0$ and we stay with with $\log(n+1)$, but because $n > 0$, we know that $\log(n+1)$ is always increasing, which means (in my opinion) that it is not converging. but the answer in my book is that it is converging. any ideas?

P.S. I just started studying it, sorry if I write nonsense..

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  • $\begingroup$ Your reasoning looks valid and convincing (except that just because $\log(n+1)$ is increasing doesn't necessarily mean that it diverges to infinity, but it happens to do so anyway). Are you sure you have copied the problem accurately? $\endgroup$ – Henning Makholm Dec 27 '15 at 16:11
  • $\begingroup$ The series does tend to infinity, as $log(n + 1) \to \infty$. Either the book is wrong, or (more likely) they include tend to infinity in the term "converge" $\endgroup$ – Milen Ivanov Dec 27 '15 at 16:12
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We have, by telescoping terms, $$ \sum_{k=1}^\infty \log\left(1+\frac{1}{k}\right)=\lim_{n \to \infty}\left( \log(n+1)-\log 1\right)=+\infty $$ your series is clearly divergent.

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  • $\begingroup$ That's exactly what I think too. Thanks Olivier! $\endgroup$ – Ilan Aizelman WS Dec 27 '15 at 16:15
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HINT:

$$\sum_{k=1}^{\infty}\ln\left(1+\frac{1}{k}\right)=\lim_{m\to\infty}\sum_{k=1}^{m}\ln\left(1+\frac{1}{k}\right)=\lim_{m\to\infty}\left[\ln((m+1)!)-\ln(m!)\right]=\lim_{m\to\infty}\ln(m+1)$$

As we know, if $m\to\infty$ than $\ln(m+1)\to\infty$

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Note that if $x\to 0$, $log(1+x)=x-o(x^2)$ therefore the series $$\sum_{k=1}^{\infty}\ln\left(1+\frac{1}{k}\right)$$ diverges. Indeed if $k\to +\infty$ $$\ln\left(1+\frac{1}{k}\right)=\frac {1}{k}-o(\frac{1}{k^2})$$ but $$\sum\frac {1}{k}$$ diverges.

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