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I have to find $$\lim _{n\to \infty }\sum _{k=2}^n\frac{1}{k\cdot \ln\left(k\right)}$$

I tried to write the general term in different ways, to apply squeeze theorem, but I don't get to a result. If possible, I would appreciate a hint using high school knowledge.

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    You should first ask yourself whether the limit converges at all. – 2017-01-16
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    **Hint:** Have you tried the integral test ? – 2017-01-16

3 Answers 3

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Notice that

$$\sum_{k=2}^n\frac1{k\ln(k)}>\int_2^n\frac1{x\ln(x)}\ dx=\ln(\log_2(n))\to+\infty$$

as $n\to\infty$.

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    I made a mistake... edited it now – 2017-01-16
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    @Liviu Ok, I've updated :-) – 2017-01-16
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Hint:

$$\frac{2^n}{2^n\log2^n}=\frac1{\log2}\cdot\frac1n$$

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Hint: Try to use Cauchy condensation test

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    Just a personal recommendation next time is that you not have the entirety of your answer sitting on a link. – 2017-01-16