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So we have this infinite series

$$\sum^{\infty}_{k=2}\frac{1}{k^2\ln(k)}$$

It is apparent that $\ln(k)<k$ for $k>2$

Therefore we can compare

$$\sum^{\infty}_{k=2}\frac{1}{k^2\ln(k)} \text{ to } \sum^{\infty}_{k=2}\frac{1}{k^2(k)}=\sum^{\infty}_{k=2}\frac{1}{k^3}$$ Which is a convergent $p$ series where $p>1$

Establishing comparison:

$$x^2\ln(x)<x^3,\:\frac{1}{x^2\ln(x)}>\frac{1}{x^3}$$

Therefore comparison test fails $a_n>b_n$ where $b_n$ converges

Limit comparison test:

$$\lim_{k\to\infty}\frac{a_n}{b_n}=\lim_{k\to\infty}\frac{\frac{1}{k^2\ln(K)}}{\frac{1}{k^3}}=\lim_{k\to\infty}\frac{k}{\ln(k)}$$

We must use L'Hopitals Rule, let $f(x)=\frac{x}{\ln(x)}$ $$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}\frac{x}{\ln(x)} = \lim_{x\to\infty} \frac{1}{1/x} \to\infty$$

$b_n$ is convergent, so the limit comparison test fails.

According to Wolfram, the ratio test fails, and $a_n$ cannot be analytically integrated (as far as I can tell). However, Wolfram states that the series IS convergent.

What test can I use? Or a trick to prove convergence would be appreciated.

Thanks Stax!

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    $\begingroup$ Just compare to $\sum\frac{1}{k^2}$. $\endgroup$
    – vadim123
    Mar 20, 2016 at 0:22
  • $\begingroup$ That is a lot easier than I imagined. What argument can be used to prove that that series is comparable? $\endgroup$
    – helpmeh
    Mar 20, 2016 at 0:23
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    $\begingroup$ Here is the argument: $\ln k > 1$. $\endgroup$
    – vadim123
    Mar 20, 2016 at 0:24
  • $\begingroup$ $\log(k)\ge 1$ for $k$ large enough $\endgroup$
    – Friedrich Philipp
    Mar 20, 2016 at 0:24
  • $\begingroup$ Thanks guys, didnt expect it to be that simple $\endgroup$
    – helpmeh
    Mar 20, 2016 at 0:25

4 Answers 4

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One may write, for $k=3,4,\ldots$, $$ \frac{1}{k^2\ln(k)}<\frac{1}{k^2} $$ and conclude.

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  • $\begingroup$ Since my series starts at k=2, can this still be used? $\endgroup$
    – helpmeh
    Mar 20, 2016 at 0:33
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    $\begingroup$ @helpmeh: Whether a series converges or diverges only depends on the "tail" behavior. The first few terms (indeed, any finite number of terms) are irrelevant. $\endgroup$
    – user169852
    Mar 20, 2016 at 0:36
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    $\begingroup$ @helpmeh You may observe that $$0<\sum_{k=2}^\infty \frac{1}{k^2\ln(k)}=\frac{1}{4\ln 2}+\sum_{k=3}^\infty \frac{1}{k^2\ln(k)}<\frac{1}{4\ln 2}+\sum_{k=3}^\infty \frac{1}{k^2}<\infty$$ giving the convergence. $\endgroup$
    – Olivier Oloa
    Mar 20, 2016 at 0:40
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By the Cauchy condensation test, its convergence is equivalent to $$ \sum_{k=1}^\infty 2^k \frac{1}{2^{2k} \log 2^k} = \sum_{k=1}^\infty \frac{1}{k 2^k \log 2} $$ Then use the root test.

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  • $\begingroup$ above my current scope of infinite series, but thank you $\endgroup$
    – helpmeh
    Mar 20, 2016 at 0:29
  • $\begingroup$ This one seems more complicated than what is needed for the occasion. $\qquad$ $\endgroup$
    – Michael Hardy
    Mar 20, 2016 at 0:32
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Not quite: $$\dfrac1{k^2\ln k}=o\Bigl(\frac1{k^2}\Bigr),$$ hence it converges.

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How about the integral test for convergence?

Note that $$\int_2^\infty \frac{\text{d}x}{x^2\ln x}$$ converges

and thus your sum converges.

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