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This question already has an answer here:

Evaluate if the following series is convergent or divergent: $\sum\limits_{n=1}^\infty \ln({1+\frac 1 {{n}}})$.

I tried to evaluate the divergence, applying the Weierstrass comparison test:

$\sum\limits_{n=1}^\infty \ln({1+\frac 1 {{n}}})>\sum\limits_{n=1}^\infty \ln({\frac 1 {{n}}})$. Since the function is decreasing then as $1+\frac{1}{n}$ is closer to $0$ then the inequality follows.

Obviously $\sum\limits_{n=1}^\infty \ln({\frac 1 {{n}}})$ diverges since $\lim_{n\to\infty}\ln({\frac 1 {{n}}})=-\infty$.

Questions:

1) Is my answer right? If not why?

2) What other kind of approach do you propose?

Thanks in advance!

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marked as duplicate by Xander Henderson, Namaste calculus Mar 31 '18 at 17:11

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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$$\sum_{n=1}^N\ln\left(1+\frac{1}{n}\right)=\ln(N+1)\to\infty\textrm{ as }N\to\infty$$

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  • $\begingroup$ How do I prove $\sum_{n=1}^N\ln\left(1+\frac{1}{n}\right)=\ln(N+1)$? $\endgroup$ – Pedro Gomes Mar 31 '18 at 17:05
  • $\begingroup$ $\sum_{n=1}^N\ln\left(1+\frac{1}{n}\right)=\ln(\frac{2}{1})+\ln(\frac{3}{2})+\ln(\frac{4}{3})+\cdots+\ln(\frac{N+1}{N})=\ln(\frac{2}{1}\cdot\frac{3}{2}\cdot\frac{4}{3}\cdots\frac{N+1}{N})=\ln(N+1)$ $\endgroup$ – CY Aries Mar 31 '18 at 17:08
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Let S be the sum of the series. The comparison you used only tells you that $S > -\infty,$ which is useless.

To determine convergence, try using the fact that $\ln a + \ln b = \ln(ab).$ Notice that $1+1/n = \frac{n+1}{n}.$

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  • $\begingroup$ I end up with $\sum_\limits{n=1}^{\infty}(\ln(n+1)-\ln(n))$.How can I prove with this expression that series diverge? Thanks for your answer! $\endgroup$ – Pedro Gomes Mar 31 '18 at 17:36
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1) Your answer is not correct. The only way you can say that $\sum_{n=1}^{\infty} a_n$ diverges because $\sum_{n=1}^{\infty} b_n$ diverges and $a_n\geq b_n$ (comparison test) is if $\sum_{n=1}^{\infty} b_n \to \infty$. For a concrete example, take $a_n=0$ and $b_n=-1$.

2) I would recommend using the limit comparison test on $\ln(1+1/n)$ and $1/n$.

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Isn't right because $\log(1/n)<0$, But you can use that $$n\to\infty\implies \log(1+1/n)\approx\frac1n.$$

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1) Well, you haven't shown that $\sum\limits_{n=1}^\infty \ln({1+\frac 1 {{n}}})$ diverges. You've show that the sum is $\ge -\infty$ but that is true of all series, right?

2) Try comparing $\ln(1+\frac{1}{n})$ to $\frac{1}{n}.$

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HINT: $$ \lim_{n\to\infty}\frac{\ln(1+(1/n))}{\frac1n}=1. $$

And your evaluation of strictly positive series by negative series from below is incorrect (useless).

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use the inequality $\text{log}(1+x) > \frac{x}{1+x}$ for any nonzero $x \geq -1$ and you get $\text{log}(1+1/n)> \frac{1}{1+ n}$. Summing over both sides over all $n \in \mathbb{N}$ yields the harmonic series on the right, which does not converge.

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Hint: Assume that $p_n\ge0$ with $p_n\to 0$ as $n\to\infty$. Then, by $$\lim_{n\to \infty}\frac{\log(1+p_n)}{p_n}=\lim_{x\to 0}\frac{\log(1+x)}{x}=1$$ and the limit comparison test you have that $\sum_{n=1}^{\infty}\log(1+p_n)$ and $\sum_{n=1}^{\infty}p_n$ are equivalent, meaning that either both converge or both diverge.

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