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$$\displaystyle \sum\limits_{}^{} \dfrac{1}{k(ln(k)^2)}$$

Integral test

$$\int_{} \frac {1}{u^2} du = \int u^{-2} du = \frac {-1}{u} = \frac{-1}{\ln k} +c$$

Jon
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  • Are you sure that it converges to zero? What does $1/\ln(1)$ equal? – CogitoErgoCogitoSum Nov 25 '14 at 22:11
  • If the integral were zero then the infinite sum would have to be zero too, which is clearly not true. – CogitoErgoCogitoSum Nov 25 '14 at 22:12
  • In the original problem this might have been my fault but the sigma notation was blank. No upper or lowerbound; just Σ – Jon Nov 25 '14 at 22:37
  • If its just a $\Sigma$ with no limits I think its fair to assume its an infinite series. But the starting point is more than likely no less than $k=2$. Otherwise the ratio is obviously undefined at the first term, then so too would the sum be. – CogitoErgoCogitoSum Nov 25 '14 at 22:41

2 Answers2

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One way is to use the Cauchy Condensation Test

See that the terms of the sequence are non negative and decreasing, so $\displaystyle \sum\limits_{k=2}^{\infty} \dfrac{1}{k(\ln^2 k)}$ converges if and only if $\displaystyle \sum\limits_{k=2}^{\infty} \dfrac{2^k}{2^k(\ln^2 2^k)} = \dfrac{1}{\ln^2 2}\sum\limits_{k=2}^{\infty} \dfrac{1}{k^2}$ converges (which is true).

Since, $\displaystyle \sum\limits_{k=2}^{\infty} \frac{1}{k^2} \le \displaystyle \sum\limits_{k=2}^{\infty} \frac{1}{k(k-1)} = \displaystyle \sum\limits_{k=2}^{\infty} \left(\frac{1}{k-1} - \frac{1}{k}\right) = 1$ (telescopes).

sciona
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3

Using the integral test as you suggested

$$\sum_{k=2}^{\infty} \frac{1}{k \ln^2k} < \int_{2}^{\infty} \frac{1}{x\ln^2x} dx = \frac{1}{\ln (2)}$$

where $\int \frac{1}{x \ln^2x} dx = - \frac{1}{\ln x} + C$

Aaron Maroja
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