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Looking at this sum:

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

I am supposed to determine weather this one diverges or conveges. At first I thought about doing the following comparison test:

$$ \sum_{n=1}^{\infty} \frac{\ln(n)}{n+1} \times \arcsin\left(\frac{1}{n}\right) > \sum_{n=1}^{\infty} \frac{1}{n+1} $$

Then, using anotheer comparison test with this sum $ \sum_{n=1}^{\infty} \frac{1}{n} $I determined that my original sum diverges.

My problem is with the $ \arcsin\left(\frac{1}{n}\right) $. I know that if it was a "regular" $\arcsin$ function, it will have a final result at infinity, but this case looks a bit different, since the whole expression goes to $0$. Does it change my answer? what should I do with that?

Thanks

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  • $\begingroup$ $\arcsin \left( \frac{1}{n} \right) \approx \frac{1}{n}$ as $n\to\infty$ $\endgroup$ – πr8 Feb 25 '16 at 12:23
  • $\begingroup$ So, does it mean that my sum shoulde converge? $\endgroup$ – MorKadosh Feb 25 '16 at 12:24
  • $\begingroup$ Yes, it should. One can show fairly quickly by means of a graph that $\arcsin x \le \frac{\pi x}{2}$ for $x \ge 0$, and then you just need to show that this sequence you use as an upper bound also has convergent sum. $\endgroup$ – πr8 Feb 25 '16 at 12:29
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Note that $\arcsin (\frac{1}{n})\sim_{+\infty} \frac{1}{n}$ Therefore using the Cauchy condensation test $$ \sum_{n=1}^{\infty} \frac{\ln(n)}{n+1} \times \arcsin\left(\frac{1}{n}\right)= \sum_{n=1}^{\infty}2^n \frac{n\ln(2)}{2^n+1} \times \arcsin\left(\frac{1}{2^n}\right) $$ but for $n\to +\infty$ the series becomes$$\frac {n\ln 2}{2^n}$$ and using the root test$$ \lim_{n\to +\infty} \left (\frac {n\ln 2}{2^n}\right )^{\frac{1}{n}}= \frac {1}{2}<1$$ therefore the series converges.

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