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$$\lim_{n\to \infty}\frac{1}{2\log2}+\frac{1}{3\log3}+...+\frac{1}{n\log n}$$

(log stands for natural logarithm). I don't have the slightest idea about how to calculate this limit.

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    Hint: Integral test or Cauchy condensation. – 2017-01-31
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    @zhw. i haven't been taught these things yet. i will look them up. thank you – 2017-01-31
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    the same idea used to prove that the harmonic serie diverges works here – 2017-01-31

2 Answers 2

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Note that $\dfrac{1}{x\ln x}$ is a decreasing function \begin{align*} \frac{1}{k\ln k} & > \frac{1}{(k+1) \ln(k+1)} \\ \frac{1}{k\ln k} & > \int_{k}^{k+1} \frac{dx}{x\ln x} \\ \sum_{k=2}^{n} \frac{1}{k\ln k} & > \int_{2}^{n+1} \frac{dx}{x\ln x} \\ &= \ln \ln (n+1)-\ln \ln 2 \end{align*}

which tends to infinity so the sum diverges.

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By Cauchy's condensation test the series $$ \sum_{n\geq 2}\frac{1}{n\log n} $$ is convergent iff the series $$ \sum_{m\geq 1}\frac{2^m}{2^m \log(2^m)} = \frac{1}{\log 2}\sum_{m\geq 1}\frac{1}{m} $$ is convergent, but the last series is the harmonic series, which is divergent (by Cauchy's condensation test again, if you like).