How to prove $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n+1}+\sqrt{n}} $ diverges using the comparison test.

Question

Question:

I'm having difficulty proving the series $$\sum_{n=1}^{\infty} \frac{1}{\sqrt{n+1}+\sqrt{n}}$$ diverges using the comparison test with with the series $\sum_{n=1}^{\infty} \frac{constant}{\sqrt{n}}$ for comparison.

Where I am at so far:

The comparison test states if 0 ≤ $a_n$ ≤ $b_n$ for all natural numbers and $\sum_{n=1}^{\infty} a_n $ diverges, then $\sum_{n=1}^{\infty} b_n $ diverges.

Let $b_n$ = $\frac{1}{\sqrt{n+1}+\sqrt{n}}$

Observe that $b_n$ = $\frac{1}{\sqrt{n+1}+\sqrt{n}}$ ≤ $\frac{1}{\sqrt{n}}$ for all natural numbers.

But the thing is, I know that if I had "$b_n$ = $\frac{1}{\sqrt{n+1}+\sqrt{n}}$ ≥ $\frac{1}{\sqrt{n}}$ for all natural numbers" I would be fine because if we let $a_n$ = $\frac{1}{\sqrt{n}}$ then I know that $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} $ diverges so by the comparison test, $\sum_{n=1}^{\infty} \frac{1}{\sqrt{n+1}+\sqrt{n}} $ diverges.

Where am I going wrong?

I have to use $\sum_{n=1}^{\infty} \frac{constant}{\sqrt{n}}$ as a comparison.

Answer 1

Note that $$ \sqrt{n+1}+\sqrt{n}\leq 2\sqrt{n+1} $$ (for $n\geq 1$ say) whence $$ \frac{1}{\sqrt{n+1}+\sqrt{n}}\geq \frac{1}{2\sqrt{n+1}} $$ from which one can conclude that $\sum \frac{1}{\sqrt{n+1}+\sqrt{n}}$ diverges using the comparison test.

An alternative which does not use the comparison test is to observe that $$ b_n=\frac{1}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}\times \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}-\sqrt{n}}=\sqrt{n+1}-\sqrt{n} $$ whence $$ \sum_{n=1}^k b_n=\sqrt{k+1}-1\to \infty $$ as $k\to \infty$.

Answer 2

Use that $$\frac{1}{\sqrt{n+1}+\sqrt{n}}\geq \frac{1}{n}$$ this is $$n^4-4n^3-2n^2+1\geq 0$$ if $n$ is large enough.

Answer 3

Use the comparison test with $$\frac1{2\sqrt{n+1}}\lt\frac1{\sqrt{n+1}+\sqrt{n}}$$ The sum over the LHS diverges by the integral test hence the sum in question also diverges.