23

$$\lim_{n\to \infty} \frac{1}{\sqrt{n}}\left(\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\cdots+\frac{1}{\sqrt{2n-1}+\sqrt{2n+1}}\right)=? $$

I tried with the squeeze theorem, though I got the upper bound, but I couldn't find the lower bound. I also tried to solve it with the order limit theorem, but without any success. I guessed the result should be $\frac{1}{\sqrt{2}}$. How can I do this?

Kushal Bhuyan
  • 7,110

5 Answers5

44

There is a telescopic sum in disguise. Since: $$\frac{1}{\sqrt{2k+1}+\sqrt{2k-1}}=\frac{1}{2}\left(\sqrt{2k+1}-\sqrt{2k-1}\right)\tag{1}$$ by summing $(1)$ for $k$ that goes from $1$ to $n$ we have: $$ \sum_{k=1}^{n}\frac{1}{\sqrt{2k+1}+\sqrt{2k-1}} = \frac{1}{2}\left(\sqrt{2n+1}-\sqrt{1}\right)\tag{2}$$ hence by multiplying both sides by $\frac{1}{\sqrt{n}}$ and taking the limit as $n\to +\infty$ we clearly have: $$ \lim_{n\to +\infty}\frac{1}{\sqrt{n}}\sum_{k=1}^{n}\frac{1}{\sqrt{2k+1}+\sqrt{2k-1}}=\color{red}{\frac{1}{\sqrt{2}}}.\tag{3}$$

Jack D'Aurizio
  • 353,855
  • 12
    (+1) I was reading the preceding answers and couldn't believe that no one saw that the series telescoped. Glad to see yours. – robjohn Sep 12 '15 at 08:59
  • 6
    This is why they always tell you in school to rationalize your denominators :) – eudes Sep 12 '15 at 17:15
13

Hint 1:
$\frac{1}{\sqrt1+\sqrt3}=\frac{\sqrt3-\sqrt1}{2}$

$\frac{1}{\sqrt3+\sqrt5}=\frac{\sqrt5-\sqrt3}{2}$

and so on, so you are left with $\frac{\sqrt{2n+1}-1}{2}$ inside the brackets,
hope you can do it from here now.

Hint 2: (see only if you need it)

After multiplying by $\frac{1}{\sqrt{n}}$ it becomes $\frac{\sqrt{2+\frac{1}{n}}-\frac{1}{\sqrt{n}}}{2}. $ Now what is $\frac 1n$ and $\frac{1}{\sqrt{n}}$ if $n \to{+ \infty}$?

Display name
  • 1,033
7

Other hint.

For $n \in \mathbb N$ and $x \in [n,n+1]$ you have $$\frac{1}{2\sqrt{2x+1}} \le \frac{1}{\sqrt{2n-1}+\sqrt{2n+1}} \le \frac{1}{2\sqrt{2x-3}}$$ which enables to compare the sum with integrals.

mathcounterexamples.net
  • 70,018
4

Notice, we have $$\lim_{n\to \infty}\frac{1}{\sqrt n}\left(\frac{1}{\sqrt 1+\sqrt 3}+\frac{1}{\sqrt 3+\sqrt 5}+\cdots +\frac{1}{\sqrt {2n-1}+\sqrt {2n+1}}\right)$$ $$=\lim_{n\to \infty}\sum_{r=1}^{n}\frac{1}{\sqrt n}\left(\frac{1}{\sqrt {2r-1}+\sqrt {2r+1}}\right)$$ $$=\lim_{n\to \infty}\sum_{r=1}^{n}\frac{1}{\sqrt n}\left(\frac{\sqrt {2r+1}-\sqrt {2r-1}}{2r+1-(2r-1)}\right)$$ $$=\frac{1}{2}\lim_{n\to \infty}\sum_{r=1}^{n}\frac{\sqrt {2r+1}-\sqrt {2r-1}}{\sqrt n}$$ $$=\frac{1}{2}\lim_{n\to \infty}\sum_{r=1}^{n}\frac{\sqrt{2r}\left(1+\frac{1}{2r}\right)^{1/2}-\sqrt{2r}\left(1-\frac{1}{2r}\right)^{1/2}}{\sqrt n}$$ Using binomial expansion of $\left(1-\frac{1}{2r}\right)^{1/2}$ & neglecting higher power terms as $\left(\frac{1}{2r}\right)^2$, $\left(\frac{1}{2r}\right)^3,\ldots $

$$=\frac{\sqrt 2}{2}\lim_{n\to \infty}\sum_{r=1}^{n}\frac{\sqrt r}{\sqrt n} \left[\left(1+\frac{1}{2}\frac{1}{2r}\right)-\left(1-\frac{1}{2}\frac{1}{2r}\right)\right]$$ $$=\frac{1}{\sqrt 2}\lim_{n\to \infty}\sum_{r=1}^{n}\frac{\sqrt r}{\sqrt n} \left[\frac{1}{2r}\right]$$ $$=\frac{1}{2\sqrt 2}\lim_{n\to \infty}\sum_{r=1}^{n}\frac{1}{\sqrt {nr}}$$ $$=\frac{1}{2\sqrt 2}\lim_{n\to \infty}\sum_{r=1}^{n}\frac{\frac{1}{n}}{\sqrt {\frac{r}{n}}}$$ $$=\frac{1}{2\sqrt 2}\int_{0}^{1}\frac{dx}{\sqrt {x}}$$ $$=\frac{1}{2\sqrt 2}[2\sqrt x]_{0}^{1}$$ $$=\frac{1}{2\sqrt 2}[2-0]$$ $$=\frac{1}{\sqrt 2}$$

Michael Hardy
  • 1
Harish Chandra Rajpoot
  • 37,450
  • 8
    Such an expensive solution for such a simple problem! – Jack D'Aurizio Sep 12 '15 at 09:04
  • 5
    @Jack: Yes and no -- it may seem gratuitous to someone just learning the material, but once this stuff becomes old-hat, it is often an easier and faster method to solve a problem than the "simple" solution. –  Sep 12 '15 at 09:41
  • 5
    It might be useful to give some explanation of why we can ignore the higher order terms of the binomial expansion. – robjohn Sep 12 '15 at 13:13
  • 2
    @robjohn: This answer is completely wrong, especially so after his edit. $\frac{1}{2r}$ is not very small as it always starts from $r = 1$ in the sum! Moreover $\sum_{k=1}^n \frac{1}{n}$ cannot be thrown away just because $\frac{1}{n}$ is so-called very small for large $n$! – user21820 Sep 12 '15 at 13:19
  • Alright, I edited to add explanation. – Harish Chandra Rajpoot Sep 12 '15 at 13:19
  • @Hurkyl: Please check before affirming that a method actually works. You cannot just perform binomial expansion of an expression within a sum and throw away terms just like that. – user21820 Sep 12 '15 at 13:23
  • 6
    @HarishChandraRajpoot: $\frac1r$ is not, of itself, small since $r$ ranges from $1$ upwards. You are relying on the fact that $\lim\limits_{n\to\infty}\sum\limits_{r=1}^n\frac{\sqrt{r}}{\sqrt{n}}\frac1{r^k}=0$ for $k\gt1$. This, at least, should be mentioned. – robjohn Sep 12 '15 at 13:30
  • @user21820: Sure you can, if you recognize that the neglected terms don't affect the result. Yes, the minor details probably should have been spelled out in this particular setting as robjohn says, but that doesn't make the method wrong. –  Sep 12 '15 at 15:19
  • @Hurkyl: If you check the intermediate edit you will see that it was indeed an incorrect method. Of course when the steps are ambiguous one can always argue that the correct analysis was intended, but almost always students do not even know the correct analysis. – user21820 Sep 12 '15 at 15:28
3

If you don't spot that the series telescopes (and this is always worth a check - usually careful inspection of the first few terms will suffice) then here is a solution using the Stolz–Cesàro theorem.

Write the desired series as a fraction:

$$S_n = \frac{\sum_{k=1}^n\frac{1}{\sqrt{2k-1}+\sqrt{2k+1}}}{\sqrt{n}} = \frac{a_n}{b_n} $$

Now $b_n = \sqrt{n}$ is strictly monotone and divergent (since it is strictly increasing and $b_n \to +\infty$). We need this condition for Stolz–Cesàro to apply.

Now consider the "fraction of differences" (does it have a more technical name?):

$$\frac{a_n - a_{n-1}}{b_n - b_{n-1}} = \frac{\left(\sqrt{2n+1}+\sqrt{2n-1}\right)^{-1}}{\sqrt{n}-\sqrt{n-1}} = \frac{\sqrt{n}+\sqrt{n-1}}{\sqrt{2n+1}+\sqrt{2n-1}}=\frac{1+\sqrt{1-\frac{1}{n}}}{\sqrt{2+\frac{1}{n}}+\sqrt{2-\frac{1}{n}}}$$

The numerator $1+\sqrt{1-\frac{1}{n}} \to 2$ and the denominator $\sqrt{2+\frac{1}{n}}+\sqrt{2-\frac{1}{n}} \to 2\sqrt{2}$ so the fraction tends to $\frac{2}{2\sqrt{2}}=\frac{1}{\sqrt{2}}$.

Then since $b_n$ was strictly monotone and divergent, and the limit of $\frac{a_n - a_{n-1}}{b_n - b_{n-1}}$ exists and equals $\frac{1}{\sqrt{2}}$, we can conclude from the Stolz–Cesàro theorem that the limit of $\frac{a_n}{b_n}$ also exists and also equals $\frac{1}{\sqrt{2}}$.

Silverfish
  • 1,620
  • 21
  • 33