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Prove $\sum_{n\ge0}\frac{\ln(n+1)}{n^2}$ Converges.

I want to know what's wrong with my proof:

By Cauchy condensation test:
$\sum_{n\ge0}\frac{\ln(n+1)}{n^2}$ converges iff $\sum_{n\ge0}2^n\frac{\ln(2^n+1)}{2^{2n}} = \sum_{n\ge0}\frac{\ln(2^n+1)}{2^n}$

Now, using the root test for the last series:

$\root {n} \of {\frac{ln(2^n+1)}{2^n}} = \frac{ln(2^n+1)}{2} = \infty$

So the conclusion is the original series diverges, I was asked to show the series converges so where is my proof failing at?

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you forgot the $n$-root of the logarithmic term – daw Jun 19 '14 at 11:47
up vote 3 down vote accepted

Can't you just use comparison? $$0 \leq \frac{\ln(n+1)}{n^2} =o\left(\frac{1}{n^{3/2}}\right)$$ As for your proof, you forgot to apply the $n$-th root to the logarithm: $$ \ln^{\frac{1}{n}}(n+1) = \exp_- \frac{\ln\ln(n+1)}{n} \xrightarrow[n\to\infty]{} e^0=1 $$

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Thank you. How did you know to choose specifically this $b_n = \frac{1}{n^{3/2}}$ – AnnieOK Jun 19 '14 at 11:52
We want to "steal" from $n^2$ a positive power of $n$ to (in the long run) crush the logarithm. Any positive power will work for the crushing. But we should not steal too much, must leave enough for the resulting series to converge. Stealing $n^{1/4}$ is fine, also $n^{1/2}$, also $n^{9/10}$. – André Nicolas Jun 19 '14 at 12:07

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