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$$\sum_{n=1}^\infty\frac1{(\sqrt n + n )\ln (\sqrt n+1)}$$

So im guessing you go $\frac{1}{f(x)}$ and $\frac{1}{g(x)}$

and see if $\frac{1}{f(x)}$ and $\frac{1}{g(x)}$ are both convergent by finding the antiderivative of both.

Is that the right way to do it? It doesn't seem to be the right way for as I go around in circles doing it this way.

Thanks for the help

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2 Answers 2

up vote 2 down vote accepted

Note that $\sqrt{n} < n$. Hence, $n+\sqrt{n} < 2n \implies \dfrac1{n+\sqrt{n}} > \dfrac1{2n}$.

Similarly, $\sqrt{n}+1 < n$ for $n>2$. Hence, $\log(\sqrt{n}+1) < \log n \implies \dfrac1{\log(\sqrt{n}+1)} > \dfrac1{\log n}$.

This gives us that $$\dfrac1{n+\sqrt{n}} \dfrac1{\log(1+\sqrt{n})} > \dfrac1{2n \log n}$$ Now use integral test to conclude that $\displaystyle \sum_{n=1}^{\infty} \dfrac1{2n \log n}$ diverges and hence $$\displaystyle \sum_{n=1}^{\infty} \dfrac1{n+\sqrt{n}} \dfrac1{\log(1+\sqrt{n})}$$ diverges.

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Ok i understand it now thank you –  user1730308 Jan 28 '13 at 2:53

You can deduce convergence or divergence by considering the behavior of the sum and for large $n$, which goes as $\frac{1}{n \log{n}}$ in this limit. By the integral test, we may deduce that a partial $N$ sum behaves as $\log{\log{N}}$ for large $N$, so this sum diverges.

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