I don't know how to prove or disprove convergence of the following. Which convergence test do I need to use?
$$\sum_{n=1}^{\infty} \frac{1}{n\ln^2n}$$
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1It converges. Integral Test (easiest) or Cauchy Condensation. By the way, the sum should not start at $1$. – André Nicolas Dec 03 '15 at 18:16
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can you write the solution? – F.leon Dec 03 '15 at 18:17
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@F.leon Try u-sub $lnn=u$ and see what happens when integrating – imranfat Dec 03 '15 at 18:19
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Someone may write a full solution soon. But if not, imranfat's hint comes close to finishing. The integral $\int_2^\infty \frac{1}{x\ln^2 x},dx$ can be evaluated explicitly using $u=\ln x$. – André Nicolas Dec 03 '15 at 18:22
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You can find many solved exampled of series in the catalog of series – Dec 03 '15 at 18:23
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By the way, for series whose general term is of the form $\frac{1}{n^\alpha \ln^\beta n}$, a general and quite simple criterion is known -- this is a "séris de Bertrand" (it's unclear to me if an English name exists, and even if this is taught as theorem in te UK/US, however. – Clement C. Dec 03 '15 at 19:31
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$$\sum_{n=2}^{\infty}\frac{1}{n \ln^2 n}<\ln2+\int_2^\infty \frac{1}{x(\ln x)^2}dx=\ln 2 - \left[1\over\ln x\right]_2^\infty=\ln 2+\frac{1}{\ln 2}$$
Kay K.
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