1
$\begingroup$

I don't know how to find the sum of $\sum\limits_{k=1}^{n}\frac{1}{\sqrt{2k}+\sqrt{2k+2}}$. After rationalization we have $\left(\frac{1}{\sqrt{2}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{6}}+...+\frac{1}{\sqrt{2n}+\sqrt{2n+2}}\right){/}\left((\sqrt{2}+\sqrt{4})(\sqrt{4}+\sqrt{6})...(\sqrt{2n}+\sqrt{2n+2})\right)=$ $(\sqrt{4}+\sqrt{6})...(\sqrt{2n}+\sqrt{2n+2})+(\sqrt{2}+\sqrt{4})...(\sqrt{2n}+\sqrt{2n+2})+...+(\sqrt{2}+\sqrt{4})...(\sqrt{2n-2}+\sqrt{2n})$

How to reduce this sum to general form?

$\endgroup$
  • 2
    $\begingroup$ $$\frac{1}{\sqrt{2k}+\sqrt{2k+2}} = \frac{\sqrt{2k}-\sqrt{2k+2}}{2}$$ which telescopes over $\sum_{k=1}^{n}$ $\endgroup$ – mattos Jan 30 '16 at 11:22
3
$\begingroup$

$$\lim\limits_{n\to\infty}\left(\frac{1}{\sqrt{n}}\sum\limits_{k=1}^{n}\frac{1}{\sqrt{2k}+\sqrt{2k+2}}\right)$$ $$=\lim\limits_{n\to\infty}\left(\frac{1}{\sqrt{n}}\sum\limits_{k=1}^{n}\frac{1}{\sqrt{2k+2}+\sqrt{2k}}\right)$$ $$=\lim\limits_{n\to\infty}\left(\frac{1}{\sqrt{n}}\sum\limits_{k=1}^{n}\frac{\sqrt{2k+2}-\sqrt{2k}}{(\sqrt{2k+2}+\sqrt{2k})(\sqrt{2k+2}-\sqrt{2k})}\right)$$ $$=\lim\limits_{n\to\infty}\left(\frac{1}{\sqrt{n}}\sum\limits_{k=1}^{n}\frac{\sqrt{2k+2}-\sqrt{2k}}{2k+2-2k}\right)$$ $$=\lim\limits_{n\to\infty}\left(\frac{1}{\sqrt{n}}\sum\limits_{k=1}^{n}\frac{\sqrt{2k+2}-\sqrt{2k}}{2}\right)$$ $$=\frac{1}{\sqrt{2}} \lim\limits_{n\to\infty}\left(\frac{1}{\sqrt{n}}\sum\limits_{k=1}^{n}(\sqrt{k+1}-\sqrt{k})\right)$$ $$=\frac{1}{\sqrt{2}} \lim\limits_{n\to\infty}\left(\frac{1}{\sqrt{n}}\cdot (\sqrt{n+1}-1)\right)$$ $$=\frac{1}{\sqrt{2}} \lim\limits_{n\to\infty}\left( \sqrt{1+\frac{1}{n}}-\frac{1}{\sqrt{n}}\right)$$ $$=\frac{1}{\sqrt{2}}$$

$\endgroup$
0
$\begingroup$

Hint rationlize you will get a telescoping series for eg after rationalizing first two terms we get $\frac{\sqrt{4}-\sqrt{2}+\sqrt{6}-\sqrt{4}}{2}=\frac{\sqrt{6}-\sqrt{2}}{2}$ on rationalizing till $n$ see which terms remain then you will get your steps

$\endgroup$
-1
$\begingroup$

Notice:

  • $$\frac{1}{\sqrt{2k}+\sqrt{2k+2}}=\frac{1}{\sqrt{k}\sqrt{2}+\sqrt{k+1}\sqrt{2}}=\frac{1}{\sqrt{2}\left(\sqrt{k}+\sqrt{k+1}\right)}$$
  • $$\sum_{k=1}^{n}\frac{1}{\sqrt{2k}+\sqrt{2k+2}}=\frac{\sqrt{n+1}-1}{\sqrt{2}}$$
  • $$\frac{1}{\sqrt{n}}\cdot\frac{\sqrt{n+1}-1}{\sqrt{2}}=\frac{\sqrt{n+1}-1}{\sqrt{2}\sqrt{n}}$$
$\endgroup$
  • $\begingroup$ Omits THE argument which makes the result holds while expanding on irrelevant steps. $\endgroup$ – Did Jan 30 '16 at 22:15

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.