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$$\sum\frac{(\ln n)^2} {n^2}$$

Any other solution for this apart from Cauchy's condensation test?

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    prove that the Limit is Zero if $n$ tends to infinity – 2017-02-08
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    Uhh... no. That is not at all sufficient to prove convergence of a series. – 2017-02-08
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    i have not seen the Sigma symbol – 2017-02-08
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    The sequence being a null sequence is a requirement for convergence but not a sufficient condition. – 2017-02-08

4 Answers 4

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Hint: For large $n$, $\ln n<\sqrt[3]n$ and thus $\frac{(\ln n)^2}{n^2}

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$f(x):=a_n$ is positive and decreasing from n=3 onward. $\int_{3}^{\infty }\frac{\ln^2(n)}{n^2} =\frac{1}{3}\left(2+\ln ^2\left(3\right)+\ln \left(9\right)\right)$ since the integral is convergent the series is too by the integral test. This is called the Maclaurin–Cauchy test. So it is not the condensation test I suppose?

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$S=\sum_{n\geq 1}\frac{\log^2(n)}{n^2}$ is convergent by the p-test, since the general term is non negative and bounded by $\frac{1}{n^{3/2}}$ for any $n$ large enough. We also have $S=\zeta''(2)$ and it is not terribly difficult to show that such number is just a bit less than $2$, by exploiting the integral representation for the Riemann $\zeta$ function, differentiation under the integral sign and the Cauchy-Schwarz inequality.

An extremely similar series, that is also an upper bound for the given series, namely $$ \sum_{n\geq 1}\frac{H_n^2}{n^2} $$ has the simple closed form $\frac{17\pi^4}{360}.$

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Cauchy condensation shows $$\sum2^n\frac{(\ln 2^n)^2} {2^{2n}}=\sum\frac{n^2(\ln 2)^2} {2^{n}}$$ then apply ratio criterion.