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

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.

• start with this: $\sqrt{n+1}+\sqrt{n} < 2\sqrt{n+1}$ Jul 29, 2019 at 14:59
• $\frac{1}{\sqrt{n+1}+\sqrt{n}}\ge\frac12 \frac{1}{\sqrt{n+1}}$ Jul 29, 2019 at 14:59

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 $$k\to \infty$$.
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.
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.