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

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?


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: $$ \lim_{n\to +\infty}\frac{1}{\sqrt{n}}\sum_{k=1}^{n}\frac{1}{\sqrt{2k+1}+\sqrt{2k-1}}=\color{red}{\frac{1}{\sqrt{2}}}.\tag{3}$$

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    $\begingroup$ (+1) I was reading the preceding answers and couldn't believe that no one saw that the series telescoped. Glad to see yours. $\endgroup$ – robjohn Sep 12 '15 at 8:59
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    $\begingroup$ This is why they always tell you in school to rationalize your denominators :) $\endgroup$ – eudes Sep 12 '15 at 17:15

Hint 1:


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 $n \to{+ \infty}$?

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    $\begingroup$ There were some little mistakes, so I corrected them. Also some formatting changes to make it more concise. Hope you don't make, but feel free to change back, if you do. $\endgroup$ – eudes Sep 12 '15 at 17:23
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    $\begingroup$ @eudes thnx for the edit $\endgroup$ – Display name Sep 12 '15 at 17:30

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.

  • $\begingroup$ How would it help to know the range of each term when what you want to do is to calculate an exact value? Could you please expand on this hint? $\endgroup$ – HelloGoodbye Sep 13 '15 at 12:03
  • $\begingroup$ Because if you compute RHS and LHS and divides by $\sqrt{n}$, both sides have the same limit. $\endgroup$ – mathcounterexamples.net Sep 13 '15 at 12:14

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 $

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

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    $\begingroup$ Such an expensive solution for such a simple problem! $\endgroup$ – Jack D'Aurizio Sep 12 '15 at 9:04
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    $\begingroup$ @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. $\endgroup$ – user14972 Sep 12 '15 at 9:41
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    $\begingroup$ It might be useful to give some explanation of why we can ignore the higher order terms of the binomial expansion. $\endgroup$ – robjohn Sep 12 '15 at 13:13
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    $\begingroup$ @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 $n$! $\endgroup$ – user21820 Sep 12 '15 at 13:19
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    $\begingroup$ @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. $\endgroup$ – robjohn Sep 12 '15 at 13:30

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 $\frac{1}{\sqrt{2}}$.

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    $\begingroup$ I have learned a new result, thank you. @silverfish $\endgroup$ – Kushal Bhuyan Sep 14 '15 at 7:12

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