3
$\begingroup$

$$\sum_{n=1}^\infty\frac{\sqrt{\vphantom{A^b}n+1}-\sqrt{n}}{n^x}$$

I multiplied by the conjugate and was able to simplify this to

$$\sum_{n=1}^\infty\frac{1}{n^x(\sqrt{\vphantom{A^b}n+1}+\sqrt{n})}$$

so the series will converge, as the values go to zero as $n$ goes to $\infty$. However, I can't figure out for which $x$ it converges. I want to say it's for all $x>0$, but I'm not sure how to justify this.

Any help appreciated!

$\endgroup$
  • 3
    $\begingroup$ if the values goes to $0$ that does not implies the series converges. $\endgroup$ – Kushal Bhuyan Jul 4 '16 at 14:14
2
$\begingroup$

Hint: Your rewriting is indeed very useful. By limit comparison or just plain comparison, your series converges iff $\sum \frac{1}{n^{x+1/2}}$ converges. Now use known results about $p$-series. The conclusion should be that we have convergence if and only if $x\gt 1/2$.

$\endgroup$
1
$\begingroup$

If $\;x>\frac12\iff x=\frac12+\epsilon\,,\,\,\epsilon>0\;$ :

$$\frac1{n^x\left(\sqrt{n+1}+\sqrt n\right)}=\frac1{n^{1/2+\epsilon}\left(\sqrt{n+1}+\sqrt n\right)}\le\frac1{n^{1/2+\epsilon}\sqrt n}=\frac1{n^{1+\epsilon}}$$

and we have convergence by comparison.

If $\;x<\frac12\;$

$$\frac1{n^x\left(\sqrt{n+1}+\sqrt n\right)}\ge\frac1{n^{1/2}\left(\sqrt{n+1}+\sqrt n\right)}\ge\frac1{2n^{1/2}(n+1)^{1/2}}\ge\frac12\frac1{\sqrt{n+1}}$$

and the series diverges, again by comparison.

$\endgroup$

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.