1
$\begingroup$

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.

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

3 Answers 3

1
$\begingroup$

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 $k\to \infty$.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.