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$$s=\sum_{n=2}^\infty \frac{1}{n (\ln n)^2}.$$

So I use the integral test and when I do so the integrand is
$$\int\frac{1}{x\;(\ln x)^2}dx =\frac{-1}{\ln x}$$ which converges to $\dfrac{1}{\ln2}$ therefore the series is also convergent.

And I need to find an integer such that $$\left\lvert s-\sum_{n=2}^\infty \frac{1}{n\;(\ln n)^2}\right\rvert < 10^{-3}.$$

How do I set this up? Is taking the integral of a series the same as summing it?

Olivier Oloa
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stackdsewew
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1 Answers1

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Hint. By using that $ x \mapsto \dfrac1{x(\ln x)^2}$ is decreasing over $[2,\infty)$, one has $$ \left|s-\sum_{n=2}^{N-1} \frac1{n(\ln n)^2}\right|=\sum_{n=N}^\infty \frac1{n(\ln n)^2}\le \frac1{N(\ln N)^2}+\int_N^\infty \frac{dx}{x(\ln x)^2}=\frac1{N(\ln N)^2}+\frac1{\ln N} $$ giving that $$ \left|s-\sum_{n=2}^{N-1} \frac1{n(\ln n)^2}\right|\leq 10^{-3} $$ for $N\geq 10^{435}$, since $435\cdot \ln(10)=1001.62$. The convergence rate is very slow.

Olivier Oloa
  • 120,989
  • Sorry where did the N come from? In the original question there was an $\infty$ in the place where your N is. – stackdsewew Jun 12 '16 at 11:11
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    Well, if $S$ is your given series then:

    $$\left | S- \sum_{n=2}^{N} \frac{1}{n \left ( \ln n \right )^2} \right |= \left | \sum_{n=2}^{\infty} \frac{1}{n \left ( \ln n \right )^2} - \sum_{n=2}^{N} \frac{1}{n \left ( \ln n \right )^2} \right | = \sum_{n=N}^{\infty} \frac{1}{n \left ( \ln n \right )^2}$$

    That $\displaystyle \sum_{n=2}^{N} \frac{1}{n \left ( \ln n \right )^2}$ is the partial sum of the series.

     – Tolaso Jun 12 '16 at 11:24