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I need on help on calculus homework.

$$\sum^{\infty}_{n=2}\frac{n\pi}{(\ln (n))^2}$$

I tried the comparison test, comparing it with:

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

but it's not getting me anywhere.

Appreciated for any help!

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try divergence test.. –  Halil Duru Mar 18 '13 at 22:58
    
Please, make titles informative. Note what I changed the question title to, so you can think of something similar next time. –  Pedro Tamaroff Mar 18 '13 at 23:21

2 Answers 2

Hint: The terms do not have limit $0$, so we cannot have convergence. Recall that if $\sum_m^\infty a_k$ exists, then $\lim_{k\to\infty} a_k=0$. (The converse fails.)

Remark: One can think of the above as a special case of Comparison. The series $1+1+1+\cdots$ obviously diverges. It is not hard to show that if $n$ is large enough (and it doesn't really have to be large) we have $\frac{ n\pi}{(\ln n)^2}\gt 1$.

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Try Condensation Test:

$$2^n\frac{2^n\pi}{\log^2(2^n)}=\frac{2^{2n}\pi}{n^2\log^2 n}$$

as the rightmost sequence doesn't even converge to zero, its series doesn't converge and thus doesn't our series, either.

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