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I need to find out for what $\alpha, \beta$ the following sum converges: $$\sum_{n=2}^\infty n^\alpha (\log n)^\beta$$

I thought I'd do that with the help of the integral criterion, that is to say I considered (after I had substituted $x = e^u$): $$\int_2^\infty e^{u(\alpha+1)} \cdot u^\beta \mathrm du$$

Then, I realised that for $\alpha + 1 > 0$, the exponential function will always dominate the monomial, so I only considered $\alpha + 1 \leq 0$. First, I let $\alpha + 1=0$, then I would be left with: $$\int_2^\infty u^\beta \mathrm du$$

I solved the integral and substituted back and ended up with: $$\lim_{b \to \infty} \frac{(\log b)^{\beta+1}}{\beta+1}$$

I figured this would only converge for $\beta+1 \leq 0$ and thought: "Ok, now I have some cases, where the sum converges: $\alpha = -1 \land \beta \leq -1$". To be sure, I checked the integral with the help of wolframalpha, and it told me the integral would not converge.

What did I do wrong? Is there an easier way to find $\alpha$ and $\beta$ so that the sum converges?

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up vote 5 down vote accepted

Since, for any $\varepsilon > 0$ fixed, $(\log n)^\beta / n^\varepsilon \to 0$ as $n \to \infty$, it is obvious that the sum converges for $\alpha < -1$ and diverges for $\alpha > -1$ (regardless of $\beta$). When $\alpha = -1$ and $\beta = - 1$ the sum diverges, as indicated by FX (hence also for all $\beta > -1$). We are left with the case $\alpha =-1$ and $\beta < -1$. Then you should compare with the integral $$ \int_2^\infty {\frac{1}{{x(\log x)^{1 + \delta } }}\,{\rm d}x} , $$ where $\delta > 0$ fixed. Make a change of variable $x' = \log x$ to find out that the integral (and hence the sum) converges.

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How's "(hence also for all $\beta\space>−1$)." clear? I also don't understand what to do with that integral in the $\beta < -1$ case. – mavavilj Sep 19 '15 at 15:54

For $\alpha = \beta = -1$, you have something like $\int \frac{1}{x \log x}$, which is $\log \log x$, which diverges. For $\beta < -1$, though, it's safe.

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Thanks, so my mistake I made was the assumption that $\lim_{b \to \infty} \frac{(\log b)^{\beta+1}}{\beta+1}$ converges for $\beta+1 \leq 0$; it should be $\beta + 1 < 0$. I went on and came to the conclusion that if $\alpha + 1 < 0$, $\beta$ won't affect the exponential function enough anymore, so if there is no equality, the integral converges for all $\beta$. Is that correct? – Huy Dec 12 '10 at 12:15

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