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I need to find if this series converges or diverges:

$$\sum_{n=1}^\infty \log\left(1+{1\over(n\sqrt{n})}\right)$$

I'm taking calculus III, so I'm allowed to use these test: nth root test, ratio test, nth term test, limit comparison test, direct comparison test, alternating series test.

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  • $\begingroup$ the series diverges $\endgroup$ – Dr. Sonnhard Graubner Feb 25 '17 at 6:17
  • $\begingroup$ @Dr.SonnhardGraubner No, the series converges. $\endgroup$ – Mark Viola Feb 25 '17 at 6:19
  • $\begingroup$ yes you have right it is the comparison test $\endgroup$ – Dr. Sonnhard Graubner Feb 25 '17 at 6:21
  • $\begingroup$ @Dr.SonnhardGraubner If the p series had a p<1 then it would diverge $\endgroup$ – Marwan N Feb 25 '17 at 6:23
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HINT:

$$0<\log(1+x)\le x$$

for $x>0$. Now use the comparison test.

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Alright so I use this $0<ln(1+{1\over(n\sqrt{n}})< 1/(n\sqrt n)$ and then I compare using Direct Comparison Test. Which gives us that the series converges since the $\sum_{n=1}^\infty1/(n\sqrt n)$ is a geometric p-series with p>1.

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  • $\begingroup$ Yes, you have it! Well done! $\endgroup$ – Mark Viola Feb 25 '17 at 6:19
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Such series is obviously less than $$ \sum_{n\geq 1}\frac{1}{n^{3/2}}=\zeta\left(\frac{3}{2}\right) $$ and obviously greater than $$ \log(2)+\log\prod_{p\in\mathcal{P}}\left(1+\frac{1}{p^{3/2}}\right)=\log\left(\frac{2\,\zeta(3/2)}{\zeta(3)}\right). $$

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