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Could you help me to understand for which $x$ this series converge $\displaystyle\sum \frac{\sqrt{n+1}-\sqrt{n}}{n^x}$?

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    So that would make x>\frac{1}{2}, given the fact that $\displaystyle\sum_{n=1}^\infty n^{-s}$ converges for s>1?2012-01-06

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Note that $\frac{\sqrt{n+1}-\sqrt{n}}{n^x}=\frac{\sqrt{n+1}-\sqrt{n}}{n^x}\cdot\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{n^x(\sqrt{n+1}+\sqrt{n})}.$

If $x>1/2$, then $\frac{\sqrt{n+1}-\sqrt{n}}{n^x}=\frac{1}{n^x(\sqrt{n+1}+\sqrt{n})}\leq\frac{1}{n^x(2\sqrt{n})}=\frac{1}{2n^{x+\frac{1}{2}}}.$ Since $\displaystyle\sum\frac{1}{2n^{x+\frac{1}{2}}}$ converges when $x>1/2$ (see p-series), by comparison test, $\displaystyle\sum\frac{\sqrt{n+1}-\sqrt{n}}{n^x}$ converges.

On the other hand, if $x\le 1/2$, $\frac{\sqrt{n+1}-\sqrt{n}}{n^x}=\frac{1}{n^x(\sqrt{n+1}+\sqrt{n})}\geq\frac{1}{n^x(\sqrt{n})}=\frac{1}{n^{x+\frac{1}{2}}}\geq\frac{1}{n}.$ Since the harmonic series diverges, by comparison test again, $\displaystyle\sum\frac{\sqrt{n+1}-\sqrt{n}}{n^x}$ diverges.

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We know that $\displaystyle\sum_{n=1}^\infty n^{-s}$ converges for $s>1$.

As pointed out by Norbert, $\sqrt{n+1}-\sqrt{n}= \frac{1}{\sqrt{n+1}+\sqrt{n}}<\frac{1}{\sqrt{n}+\sqrt{n}}$.

Hence, $\displaystyle\sum_{n=1}^\infty \frac{\sqrt{n+1}-\sqrt{n}}{n^x}<\displaystyle\sum_{n=1}^\infty \frac{1}{2\times{n^{\frac{1}{2}+x}}}$.

hence, for convergence, we must have the exponent of n $>1$. Hence, we must have $x>\frac{1}{2}$.

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    For $x=1/2$, it becomes Harmonic series, which we know to be divergent. For $x\le (1/2)$, by comparison test, we can see that the series diverges since Harmonic series diverges.2012-01-06