# Convergence of series sqrt(n+1)-sqrt(n-1)

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• Oct 20th 2009, 09:10 AM
calum
Convergence of series sqrt(n+1)-sqrt(n-1)
$\displaystyle \sum_{n=1}^{\infty} (\sqrt{n+1}-\sqrt{n-1})$

I need to determine whether the series converges or diverges using either the integral test, d'Alembert's ratio test or the comparison test.

So far, I can determine that the series diverges by finding a rearrangement using a finite series and then finding the limit as n approaches infinity:

$\displaystyle S_n = \sum_{k=1}^{n} (\sqrt{n+1}-\sqrt{n-1})$

$\displaystyle = (\sqrt{2}-\sqrt{0})+(\sqrt{3}-\sqrt{1})+(\sqrt{4}-\sqrt{2})+...+(\sqrt{n}-\sqrt{n-2})+(\sqrt{n+1}-\sqrt{n-1})$

$\displaystyle = \sqrt{n}+\sqrt{n+1}-1$

Hence:

$\displaystyle \sum_{n=1}^{\infty} (\sqrt{n+1}-\sqrt{n-1}) = \lim_{n \to \infty} S_n = \lim_{n \to \infty} \sqrt{n}+\sqrt{n+1}-1 = \infty$ and so the series diverges.

However, I have not used either of the three tests mentioned. Is there a way to determine convergence/divergence using those tests?

I have ascertained that for higher powers of n (p>2) of the same series, the series would converge (using the comparison test):

$\displaystyle \sum_{n=1}^{\infty} (\sqrt{n^p+1}-\sqrt{n^p-1}) = \sum_{n=1}^{\infty} \frac{2}{\sqrt{n^p+1}+\sqrt{n^p-1}} < \sum_{n=1}^{\infty} \frac{1}{\sqrt{n^p+1}}$ $\displaystyle < \sum_{n=1}^{\infty} \frac{1}{\sqrt{n^p}} = \sum_{n=1}^{\infty} \frac{1}{n^\frac{p}{2}}$

since we know that this last series (a p-series) converges for $\displaystyle \frac{p}{2} > 1$ and therefore since all the other series are lower in value and are positive series, they must also converge.

But for the case of $\displaystyle \frac{p}{2} = 1$ as with this problem, the last series would diverge, and showing that a series is less than a diverging series proves nothing.

Thanks for your help! (Happy)
• Oct 20th 2009, 11:40 AM
Krizalid
for $\displaystyle n\ge1$ is $\displaystyle \frac{2}{\sqrt{n+1}-\sqrt{n-1}}>\frac{1}{\sqrt{n+1}}>\frac{1}{2n}.$
• Oct 20th 2009, 08:41 PM
calum
Thanks heaps!

I never thought of that step: $\displaystyle \frac{1}{\sqrt{n+1}} > \frac{1}{2n}$

Actually the series simplifies to $\displaystyle \frac{2}{\sqrt{n+1}+\sqrt{n-1}}$ (note the plus sign), but I know what to do now:

$\displaystyle \frac{2}{\sqrt{n+1}+\sqrt{n-1}} > \frac{2}{2\sqrt{n+1}} = \frac{1}{\sqrt{n+1}} > \frac{1}{2n}$