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For what values of $\alpha >0$ does the series $$\sum_{k=1}^{\infty} \frac{1}{(k+1)\ln(k+1)^\alpha}$$ converge?

In the case that $\alpha = 1$, we have the $$\sum_{k=1}^{\infty} \frac{1}{(k+1)\ln(k+1)}$$ Now we consider the function $$ f(x) = \frac{1}{(x+1)\ln(x+1)}$$ for $f:[1, \infty)\rightarrow \mathbb{R}$. This function is monotonically decreasing and continuous, however, for each index $n$, $$ \int_{1}^{n} f(x)dx = \int_{1}^{n} \frac{1}{(x+1)\ln(x+1)} dx = \ln[\ln(n+1)]-\ln(\ln2)$$ by the First Fundamental Theorem (Integrating Derivatives). This sequence $\ln[\ln(n+1)]-\ln(\ln2)$ is not bounded, and therefore by the integral test, the series $$\sum_{k=1}^{\infty} \frac{1}{(k+1)\ln(k+1)}$$ diverges. Now we will have two more cases where $0<\alpha<1$ and where $\alpha>1$. This is where I get stuck. The ratio and root test are inconclusive from my attempts of using them. I would like to use the comparison test but not sure what I would compare this series too. I was wondering if anyone could please help me out.

RobPratt
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thomasbdc
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  • Any non-zero $\alpha$ makes the series converging. – Ng Chung Tak Dec 04 '20 at 01:02
  • Should those $k$ be $n$ ? – PM 2Ring Dec 04 '20 at 01:02
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    Perhaps surprisingly, the integral test works just fine for other values of $\alpha$ as well! Try the substitution $u=\ln(x+1)$ in the resulting integral. – Greg Martin Dec 04 '20 at 01:24
  • I was doing a bit of research on this problem that I am working on and others seem to conclude that the series diverges for $\alpha<1$ and converges for $\alpha>1$. Why does it work for all $\alpha>0$? – thomasbdc Dec 04 '20 at 01:48
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    Linking https://math.stackexchange.com/questions/2644338/ and the general case https://math.stackexchange.com/questions/267697/ for reference. The integral test works for all $\alpha > 0$ and will show that the series converge for $\alpha > 1$ and diverge for $\alpha \le 1$. – player3236 Dec 04 '20 at 03:00
  • It does not converge for $\alpha > 0$. What is needed is $\alpha > 1$. – marty cohen Dec 04 '20 at 03:56
  • @thomasbdc Did you make typo? $$\ln^a x \ne \ln x^a$$ – Ng Chung Tak Dec 04 '20 at 13:40

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You already resolved the case when $\alpha = 1$. So if $\alpha \neq 1$, let $u_k=\frac 1 {1-\alpha}\ln^{1-\alpha}k$. Then $$\begin{split} u_{k+1}-u_k &=\frac 1{1-\alpha}\left [\left( \ln k + \ln\left(1+\frac 1 k\right)\right)^{1-\alpha}-\ln^{1-\alpha}k\right]\\ &= \frac 1{1-\alpha}\left [\ln^{1-\alpha}(k)\left( 1 + \frac{\frac 1 k +\mathcal O\left(\frac 1 {k^2}\right)}{\ln k})\right)^{1-\alpha}-\ln^{1-\alpha}k\right]\\ &=\frac 1 {k\ln^\alpha k} + \mathcal O\left(\frac 1 {k^2\ln^\alpha k}\right)\\ \end{split}$$ Because $\sum_{k>2}\frac 1 {k^2\ln^\alpha k}$ always converges, $\sum_{k>2}\frac 1 {k\ln^\alpha k}$ converges iff $\{u_k\}$ converges. That is, if and only if $\alpha > 1$.

Stefan Lafon
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