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How would you prove convergence/divergence of the following series?

$$ \sum_{n\geq 2}\left( \dfrac{\ln(1+n)}{\ln(n)}-1\right)$$

I'm interested in more ways of proving convergence/divergence for this series.

My thoughts

$$\dfrac{\ln(1+n)}{\ln(n)}=\frac{\ln(n(1+\dfrac{1}{n})}{\ln(n)} =\frac{\ln(n)+\ln(1+\frac{1}{n})}{\ln(n)} =1+\frac{\ln(1+\frac{1}{n})}{\ln(n)}$$

then

$$\dfrac{\ln(1+n)}{\ln(n)}-1=\frac{\ln(1+\frac{1}{n})}{\ln(n)}$$

note that $\ln(1+\frac 1n)=\frac 1n+o(\frac 1n)$ then $\ln(1+\frac 1n)\sim \frac 1n$ thus $u_n-1\sim \frac 1{n\ln(n)}$

or the serie $\dfrac{1}{n\ln(n)}$ divergent by Bertrand's test

the sum up $ \sum_{n\geq 2} \dfrac{\ln(1+n)}{\ln(n)}-1$ divergent

  • Is my proof correct
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    $\begingroup$ Looks good. I don't see a mistake. $\endgroup$ – zhoraster Sep 21 '15 at 19:46
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    $\begingroup$ It depends on one's degree of fussiness. I am content, the key idea is well described. One might want to do a formal Limit Comparison with $\sum_2 \frac{1}{n\ln n}$. $\endgroup$ – André Nicolas Sep 21 '15 at 19:50
  • $\begingroup$ For clarification: Is the summation to be interpreted as $-1+\sum\frac{\ln(1+n)}{\ln n}$ or $\sum\left(\frac{\ln(1+n)}{\ln n}-1\right)$? The answers suggest the latter. $\endgroup$ – user170231 Sep 21 '15 at 21:12
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    $\begingroup$ $$\sum\left(\frac{\ln(1+n)}{\ln n}-1\right)$$ $\endgroup$ – Educ Sep 21 '15 at 21:18
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One other way to see this is to note that $$\frac{\ln(n+1)-\ln n}{\ln n}\ge \int_{\ln n}^{\ln(n+1)}\frac{dt}{t}$$ hence $$\sum_{n=2}^m\frac{\ln(n+1)-\ln n}{\ln n}\ge\int_{\ln 2}^{\ln(m+1)}\frac{dt}{t} >\ln(\ln(m+1))$$ and the conclusion follows.

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Also note, using the simple fact that $\lim_{n\to\infty} \left(1+1/n\right)^n =e$, that when $n$ large, we have

$$\dfrac{\ln(1+1/n)}{\ln(n)}=\dfrac{\ln(1+1/n)^n}{n\ln(n)}\approx \frac{1}{n \ln(n)}$$ where for getting the equality I mutiplied both numerator and denominator by $n$.

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    $\begingroup$ @ it's good idea $\endgroup$ – Educ Sep 21 '15 at 21:23
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A faster way of getting to the same series to apply the Limit Comparison Test uses the mean value theorem: $$\ln(1+n) - \ln(n) \approx \frac{1}{n}$$ $$\frac{\ln(1+n)}{\ln(n)} - 1 \approx \frac{1}{n\ln n}.$$

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  • $\begingroup$ I tired to use mean value theorem to gdt $\ln(1+n) - \ln(n) \approx \frac{1}{n}$ i took $$f(n)=\ln(n)\quad \exists n \text{ such that } \dfrac{f(n+1)-f(n)}{1}=f'(n)$$ could u add some details $\endgroup$ – Educ Sep 21 '15 at 21:30
  • $\begingroup$ Perhaps take a derivative? :~) $\endgroup$ – user217285 Sep 22 '15 at 2:06
  • $\begingroup$ would u add some clarification please $f'(n)=\dfrac{1}{n}$ $\endgroup$ – Educ Sep 22 '15 at 9:21
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There have already been several solid ways forward presented. Here we use standard inequalities only.

To that end, we have the inequalities

$$\frac1{2n}<\frac{1}{n+1}\le\log\left(1+\frac1n\right) \tag 1$$

valid for $n\ge 2$.

Therefore, for $n\ge 2$, we have that

$$\frac{\log (n+1)-\log n}{\log n}\ge \frac{1}{(n+1)\log n} \ge \frac{1}{2n\log n} \tag 3$$

Inasmuch as the series on the right-hand side of $(3)$ diverges by the integral test, then the original series also diverges. And we are done.

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  • $\begingroup$ Thank u just give me time to understand ur way, i'm kind of asleep so if i didn't do tonight may be tomorow $\endgroup$ – Educ Sep 21 '15 at 22:45
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    $\begingroup$ @Educ No worry. I just want to help. ;-)) $\endgroup$ – Mark Viola Sep 21 '15 at 22:46
  • $\begingroup$ i know and Thank you so much for ur help $\endgroup$ – Educ Sep 21 '15 at 22:46
  • $\begingroup$ Please if u could provide detailed solution here math.stackexchange.com/questions/1445661/… $\endgroup$ – Educ Sep 21 '15 at 22:48
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    $\begingroup$ @educ Sure. I posted an answer for that question. Please let me know how I can improve my answer. I really want to give you the best answer I can. $\endgroup$ – Mark Viola Sep 22 '15 at 2:09
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Note that \begin{align} \frac{\log(1+x)-\log(x)}{\log(x)} \end{align} is monotonically decreasing and positive when $x>1$. Applying Cauchy's condensation test, it suffices to look at \begin{align} \sum^\infty_{k=1}2^k\left( \frac{\log\left(1+\frac{1}{2^k}\right)}{k\log 2}\right). \end{align} Next, note that \begin{align} \frac{1}{2}x\leq \log(1+x) \end{align} for $0\leq x\leq 1$, then we see that \begin{align} \sum^\infty_{k=1}2^k\left( \frac{\log\left(1+\frac{1}{2^k}\right)}{k\log 2}\right) \geq \frac{1}{2\log 2}\sum^\infty_{k=1} \frac{1}{k}=\infty. \end{align} Hence it follows that the original series also diverges.

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