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How to determine if the following series are convergent or divergent? I'm supposed to use here the limit comparison test, but I don't know how to choose the second series. $$\sum_{k=1}^\infty \ln(1+ \sqrt{\frac 2k})$$ $$\sum_{k=1}^\infty\displaystyle \sqrt[k]{e}\sin(\frac{\pi}{k}).$$

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  • $\begingroup$ For the second one, can't we use a variation of the harmonic series? $\endgroup$ – Aranya Lahiri Nov 20 '13 at 19:45
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A related problem. Since

$$ \lim_{k\to \infty }\frac{\ln(1+\sqrt{2/k})}{\sqrt{2/k}}=1, $$

then the series diverges by the fact:

Suppose $\sum_{n} a_n$ and $\sum_n b_n $ are series with positive terms, then

if $\lim_{n\to \infty} \frac{a_n}{b_n}=c>0$, then either both series converge or diverge.

Note: We used the Taylor series

$$ \ln(1+t)=t+O(t^2)\implies \ln(1+t)\sim t. $$

$$e^t\sin(\pi t)= \pi t+O(t^2)\implies e^t\sin(\pi t) \sim \pi t.$$

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