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I'm trying to show convergence of the following:

$$ \lim\limits_{n \to \infty} \sum_{k=1}^{n} \left(\frac{1}{k \cdot \log(n)}\right) $$

How shall I proceed?

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    $\begingroup$ You might look up Euler's constant. Also, pull the $\log(n)$ out of the sum. $\endgroup$ – Ted Shifrin Jul 21 '18 at 1:00
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Hint: Draw a graph of $f(x) = \dfrac{1}{x}$ over $[1,n]$,and see that: $\dfrac{1}{2}+\dfrac{1}{3}+\cdots +\dfrac{1}{n}<\displaystyle \int_{1}^n \dfrac{1}{x}dx = \ln n < 1+\dfrac{1}{2}+\cdots +\dfrac{1}{n-1}$

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$$\frac1{\log n}\sum^n_{k=1}\frac1k=\frac{H_n}{\log n}\sim\frac{\gamma+\log n}{\log n}\to 1$$ as $n\to\infty$.

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Hint

$$S_n=\sum_{k=1}^{n} \frac{1}{k \, \log(n)}=\frac{1}{ \log(n)}\sum_{k=1}^{n} \frac{1}{k}=\frac{H_n}{ \log(n)}$$ where appears the harmonic number.

Use the asymptotics $$H_n=\gamma +\log \left({n}\right)+\frac{1}{2 n}+O\left(\frac{1}{n^2}\right)$$

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Since $\log(n)$ can be viewed as a number in the sum:

$$ \lim\limits_{n \to \infty} \sum_{k=1}^{n} \left(\frac{1}{k \cdot \log(n)}\right) = \lim\limits_{n \to \infty} \frac{1}{\log(n)} \sum_{k=1}^{n} \left(\frac{1}{k }\right) = \lim\limits_{n \to \infty} \frac{\sum_{k=1}^{n} \left(\frac{1}{k }\right)}{\log(n)} \cdot $$ since $$\lim\limits_{n \to \infty} \sum_{k=1}^{n} \left(\frac{1}{k }\right)$$ diverges so look at this, you can apply L'Hopital Rule (that will be your work).

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  • $\begingroup$ How do you use L'hopitale rule for this question ? if possible, show it here so OP can see. $\endgroup$ – DeepSea Jul 21 '18 at 2:22
  • $\begingroup$ sciweavers.org/upload/Tex2Img_1532140298/render.png I don't know if this works well in this case but it seemed reasonable $\endgroup$ – Verónica Jul 21 '18 at 2:36
  • $\begingroup$ Ok, now I tried to figure out this like: lets think this as an integral, if the integral converges then, the sum converges. lets think in this integral: www4d.wolframalpha.com/Calculate/MSP/… it converges to 1, what happens if you do the limit of 1 tending to infinity? Of course! is 1 :) $\endgroup$ – Verónica Jul 21 '18 at 2:59
  • $\begingroup$ You can check that here: wolframalpha.com/input/?i=integral++from+1+to+n+1%2F((log(n)*x) and look at that primitive, its very important $\endgroup$ – Verónica Jul 21 '18 at 3:05
  • $\begingroup$ I would be grateful if you tell me if you expected that kind of explanation (thinking in integrals) because we could find another if it didn't (like other did in the previous comments with the harmonic function) $\endgroup$ – Verónica Jul 21 '18 at 3:15

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