Convergence or divergence of $ \sum_{n=1}^{\infty}\frac{\sqrt{n+2}-\sqrt{n}}{n^{3/2}} $: How to argue?

Question

Does the series $$ \sum_{n=1}^{\infty}\frac{\sqrt{n+2}-\sqrt{n}}{n^{3/2}} $$ converge or diverge?

My attempt was to write the series as $$ \sum_{n=1}^{\infty}\frac{\sqrt{n+2}-\sqrt{n}}{n^{3/2}}=\sum_{n=1}^{\infty}\frac{\sqrt{n+2}}{n^{3/2}}-\sum_{n=1}^{\infty}\frac{1}{n} $$

The first series can be estimated from below by the harmonic series: $$ \sum_{n=1}^{\infty}\frac{\sqrt{n+2}}{n^{3/2}}=\sum_{n=1}^{\infty}\frac{(n+2)^{1/2}}{n^{1/2}\cdot n}\geqslant\sum_{n=1}^\infty \frac{1}{n}=\infty $$ and hence diverges.

The second series is the harmonic series and hence diverges.

Now, I am not sure what the whole thing does.

Answer 1

Your attempt is wrong because you can't separate the series into a difference of divergent series. You got an indeterminate form $\infty - \infty$.

You could try to rationalize the numerator: $$(\sqrt{n+2}-\sqrt{n})\frac{\sqrt{n+2}+\sqrt{n}}{\sqrt{n+2}+\sqrt{n}}=\frac{2}{\sqrt{n+2}+\sqrt{n}}$$ Then the series becomes

$$\sum_{n=1}^{\infty}\frac{2}{(\sqrt{n+2}+\sqrt{n})n^{3/2}}$$

You can use comparison test to conclude.

Answer 2

HINT

We can more effectively use that

$$\sqrt{n+2}-\sqrt{n}=\frac{2}{\sqrt{n+2}+\sqrt{n}}$$

Answer 3

Note that $\sqrt{n+2}-\sqrt n\to 0$ as $n\to\infty$, so $\;\dfrac{\sqrt{n+2}-\sqrt{n}}{n^{3/2}}=o\biggl(\dfrac 1{n^{3/2}}\biggr)$, and the latter series converges.

Answer 4

You can use the estimate: $$\frac{\sqrt{n+2}-\sqrt{n}}{n^{3/2}}<\frac1{n^2} \iff \sqrt{n}(\sqrt{n+2}-\sqrt{n})<1 \iff \\ \sqrt{n(n+2)}<n+1 \iff n^2+2n<n^2+2n+1.$$

Answer 5

Rewriting :

$2=(n+2) -n=$

$(\sqrt{n+2}-√n)(\sqrt{n+2}+√n) \gt$

$\sqrt{n+2}-√n$, since

$\sqrt{n+2}+√n >1$.

Hence

$\dfrac{\sqrt{n+2}-√n}{n^{3/2}} \lt \dfrac{2}{n^{3/2}}.$

Use comparison test.