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For the series $$\sum_{k=2}^{\infty} {1\over k(\ln k)^{1\over 2}}$$

$$\sum_{k=2}^{\infty} {1\over k(\ln k)^{2}}$$ How would I prove that the first sequence diverges and the second converges. I tried to use partial sums but I don't know how to write the equations in the form needed to use the formula. Thanks.

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  • $\begingroup$ You can try using the integral test and then using the result about $p$-series $\endgroup$ – Osama Ghani Dec 1 '17 at 3:21
  • $\begingroup$ By the way, just for the sake of future reference: what you are looking at are specific case of Bertrand series, i.e. series of the form $\sum_{n} \frac{1}{n^a \ln^b n}$ (where $a,b$ are two constants). The general result states that such a Bertrand series converges if, and only if, (i) $a>1$ or (ii) $a=1$ and $b>1$. $\endgroup$ – Clement C. Dec 1 '17 at 3:27
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Integral test: $\displaystyle\int_{2}^{\infty}\dfrac{1}{x(\log x)^{1/2}}dx=2\lim_{M\rightarrow\infty}(\log M)^{1/2}-2(\log 2)^{1/2}=\infty$.

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HINT : consider $\int_{2}^{\infty} \frac{dx}{x\log^{p}(x)}$ , where $p = 2$ or $p = 0.5$

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By the Cauchy Condensation test, the sum $$ \sum_{n=2}^\infty a_n $$ converges if and only if $$ \sum_{k=1}^{\infty}2^ka_{2^k} $$ converges. For you first series, this means we may examine the sum of $\frac{2^k}{2^kk^{1/2}\ln 2}=\frac{1}{\ln 2k^{1/2}}$ converges, which it certainly does not. Do the same procedure for the second.

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