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Find $$\lim_{n \to \infty}\frac{1}{\sqrt{n}}\left[\frac{1}{\sqrt{2}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{6}}+\cdots+\frac{1}{\sqrt{2n}+\sqrt{2n+2}}\right]$$

MY TRY: $$ \begin{align} \lim_{n \to \infty} &\frac{1}{\sqrt{n}} \biggl[ \frac{1}{\sqrt{2}+\sqrt{4}} + \frac{1}{\sqrt{4}+\sqrt{6}} + \cdots + \frac{1}{\sqrt{2n}+\sqrt{2n+2}} \biggr] \\ &= \lim_{n \to \infty} \frac{\sqrt{n}}{n} \biggl[ \frac{1}{\sqrt{2}+\sqrt{4}} + \frac{1}{\sqrt{4}+\sqrt{6}} + \cdots + \frac{1}{\sqrt{2n}+\sqrt{2n+2}} \biggr] \end{align} $$ Now using Cauchy first thm of limits $$ a_n = \frac{\sqrt{n}}{\sqrt{2n}+\sqrt{2n+1}} $$

The answer should be $\frac{1}{2\sqrt{2}}$.

But the answer is $1/\sqrt{2}$.

Mark Viola
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Rayees Ahmad
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    You should edit the question to make it readable. – Umberto P. May 05 '16 at 19:17
  • Note: I reformatted the first expression, gave up on the second. Please check the first part to make sure I got it right. – lulu May 05 '16 at 19:19
  • The problem with your approach is that you failed to account for the number of terms increasing with each $n$. Ignoring the last term that you handled, each other term individually goes to $0$. But since the number of them increases, this isn't sufficient to say that the sum goes to $0$. – Paul Sinclair May 05 '16 at 22:10

3 Answers3

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$$\frac{1}{\sqrt{2k}+\sqrt{2k+2}}=\frac{\sqrt{2k+2}-\sqrt{2k}}{2}$$

The sum telescopes to $\displaystyle \frac{1}{\sqrt n} \frac {\sqrt{2n+2} - \sqrt 2 }{2}$ which converges to $1/\sqrt{2}$

If you miss this, you can still use integrals to get bounds on $\sum_{k=1}^n \frac{1}{\sqrt{2k}+\sqrt{2k+2}}$

Gabriel Romon
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If one attempts to use Cauchy's First Limit Theorem, then we start with $$\begin{align} \lim_{n\to \infty}\frac{1}{\sqrt{n}}\sum_{k=1}^n\frac{1}{\sqrt{2k}+\sqrt{2k+2}}&=\lim_{n\to \infty}\sqrt{n}\frac1n\sum_{k=1}^n\frac{1}{\sqrt{2k}+\sqrt{2k+2}} \tag 1\\\\ \end{align}$$

The Theorem reveals that since $\lim_{n\to \infty}\frac{1}{\sqrt{2n}+\sqrt{2n+2}}=0$, then $\lim_{n\to \infty}\frac1n\sum_{k=1}^n\frac{1}{\sqrt{2k}+\sqrt{2k+2}}=0$

Therefore, the limit on the right-hand side of $(1)$ is of indeterminate form (i.e., $\infty \times 0$) and we need to pursue its evaluation judiciously.

We revert to the left-hand side of $(1)$ and note that both $\sqrt{n}$ and $\sum_{k=1}^n\frac{1}{\sqrt{2k}+\sqrt{2k+2}}$ are monotonically increasing, divergent sequences. We can invoke, therefore, the Stolz-Cesaro Theorem and write

$$\begin{align} \lim_{n\to \infty}\frac{1}{\sqrt{n}}\sum_{k=1}^n\frac{1}{\sqrt{2k}+\sqrt{2k+2}}&=\lim_{n\to \infty}\frac{\sum_{k=1}^{n+1}\frac{1}{\sqrt{2k}+\sqrt{2k+2}}-\sum_{k=1}^{n}\frac{1}{\sqrt{2k}+\sqrt{2k+2}}}{\sqrt{n+1}-\sqrt{n}} \\\\ &=\lim_{n\to \infty}\frac{\frac{1}{\sqrt{2n+2}+\sqrt{2n+4}}}{\sqrt{n+1}-\sqrt{n}}\\\\ &=\lim_{n\to \infty}\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{2n+2}+\sqrt{2n+4}}\\\\ &=\frac{1}{\sqrt{2}} \end{align}$$

as was to be shown!

Mark Viola
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This is a classic partial sums of integrals problem. Your expression is: $$\frac{1}{\sqrt{n}}\sum\limits_{k=1}^n \frac{1}{\sqrt{2k}+\sqrt{2k+2}}$$

write as $$\frac{1}{n}\sum\limits_{k=1}^n \frac{1}{\sqrt{2\frac{k}{n}}+\sqrt{2\frac{k}{n}+2}}$$

then the limit is the same as the integral $$\int_0^1\frac{1}{\sqrt{2x}+\sqrt{2x+2}}dx= \int_0^1\frac{\sqrt{2x}-\sqrt{2x+2}}{-2}dx $$

Rene Schipperus
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