# Convergence of $\sum_{n=1}^\infty \frac{\sqrt{n+1}-\sqrt n}{n^p}$ for $p \ge 0$

Fix $p \ge 0$. I'm having trouble determining whether or not $$\sum_{n=1}^\infty \frac{\sqrt{n+1}-\sqrt n}{n^p}$$ converges.

Clearly, the limit criterion is satisfied and I have tried to apply the ratio test to no success (I receive $|a_{n+1}/a_n| = 1$ and thus the test is inconclusive. I don't think the root test nor integral test would help since it is almost impossible to integrate or take the $n$ root of the summand.

Thus, I believe the key to solving this problem is the comparison test. I however cannot think of a series that I may use for comparison. Does anybody have a suggestion?

• What if you knew that $\sqrt{n+1}-\sqrt{n}=O(n^{-1/2})$? – uniquesolution Nov 6 '15 at 9:31
• Well, $\sqrt{n+1}-\sqrt n\approx{1\over2\sqrt n}$ – Ivan Neretin Nov 6 '15 at 9:31
• @IvanNeretin Yes I did already try that I was able to compare the series to $\sum \frac{1}{2n^{p+1/2}}$ but this will only converge for $p > 1/2$. I am looking for convergence when $p \ge 0$ in general. – MathMajor Nov 6 '15 at 9:33

## 1 Answer

$\frac{\sqrt{n+1}-\sqrt n}{n^p} = \frac{1}{n^p(\sqrt{n+1}+\sqrt{n})},$ and the latter is equivalent to $\frac1{2n^{p+1/2}}$ You just have to compare with Riemann series.

Without $\sim$:

Let $a_n = \frac{\sqrt{n+1}-\sqrt n}{n^p}$, and $b_n = \frac1{2n^{p+1/2}}$. Then $\frac{a_n}{b_n} \xrightarrow[n\to\infty]{} 1$, ie $a_n = O(b_n)$. Thus, if $p>1/2$, as $\Sigma b_n$ converges, then $\Sigma a_n$ converges.

Notice also that $\frac{b_n}{a_n} \xrightarrow[n\to\infty]{} 1$, ie $b_n = O(a_n)$. The reciprocal of the comparison test asserts that if $\Sigma b_n$ diverges, then $\Sigma a_n$ diverges.

Thus, if $p\leq \frac12$, as $\Sigma b_n$ diverges, then $\Sigma a_n$ diverges too.

• Please see my comment to @IvanNeretin. Here $p \ge 0$. And clearly if $p = 1/2$ the harmonic series $\sum \frac{1}{2n^{1/2+1/2}} = \sum \frac{1}{2n}$ does not converge. – MathMajor Nov 6 '15 at 9:34
• Sure. The convergence or not of your serie depends on the value of $p$. If $p\leq\frac12$, then it diverges, and if $p>\frac12$, then it converges. – Kevin Quirin Nov 6 '15 at 9:35
• Divergence of the strictly greater series does not imply divergence of the original series. – MathMajor Nov 6 '15 at 9:36
• The comparison test says : If $a_n \sim b_n$, then $\Sigma a_n$ converges iff $\Sigma b_n$ converges. Here, $a_n$ is $\frac{\sqrt{n+1}-\sqrt n}{n^p}$ and $b_n$ is $\frac1{2n^{p+1/2}}$. – Kevin Quirin Nov 6 '15 at 9:40
• Sorry we have not used the $\sim$ notation. Can this be explained without asymptotics? – MathMajor Nov 6 '15 at 9:42