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I'm trying to show whether or not

$$ \sum_{n=1}^\infty \ln \left(n \arctan \left(\frac{1}{n}\right)\right) $$

converges or diverges. I cannot see a way into this problem that concludes it converges, which is what Wolfram concluded. Every approach I've tried gets that it diverges. Even Raabe's test. Any help?

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Hint: Expand $\arctan u$ into its Taylor series, using only the first two terms $($it should suffice$)$, then use the approximation for $\ln(1+a)$ when $a\to0$.

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  • $\begingroup$ Perfect, I got it! $\endgroup$ – Marchi Feb 28 '15 at 1:18
  • $\begingroup$ It converges to about $−0.44597129538707248.$ $\endgroup$ – Lucian Feb 28 '15 at 1:20
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If you follow Lucian's recommendation and use more term for Taylor series of $\tan ^{-1}(u)$ and for $\log(1+a)$, you should arrive to something like $$\log \left(n \arctan \left(\frac{1}{n}\right)\right)=-\frac{1}{3 n^2}+\frac{13}{90 n^4}-\frac{251}{2835 n^6}+\frac{3551}{56700 n^8}+O\left(\left(\frac{1}{n}\right)^{10}\right)$$ and you know that $\sum_{i=1}^{\infty}\frac{1}{n^2}=\frac{\pi ^2}{6}$,$\sum_{i=1}^{\infty}\frac{1}{n^4}=\frac{\pi ^4}{90}$,$\sum_{i=1}^{\infty}\frac{1}{n^6}=\frac{\pi ^6}{945}$,$\sum_{i=1}^{\infty}\frac{1}{n^8}=\frac{\pi ^8}{9450}$.

So, for $$S=\sum_{n=1}^\infty \ln \left(n \arctan \left(\frac{1}{n}\right)\right)$$ using one term $S\approx -0.548311$, using two terms $S\approx -0.391976$, using three terms $S\approx -0.482047$, using four terms $S\approx -0.419164$, an infinite number of terms leading to the value Lucian gave.

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