# Evaluate limit of infinite series

So I have a homework question. (I'm not sure how you guys are using all the symbols so it will be ugly hand typed)

So my question is, how would you evaluate this:

$$\lim_{n\to\infty}\frac1{\sqrt n\sqrt{n+1}}+\frac1{\sqrt n\sqrt{n+2}}+\ldots+\frac1{\sqrt n\sqrt{n+n}}$$

I've tried to convert it into a definite integral, but I'm getting pretty confused on doing that. I have read around, mostly this link, but I'm still confused on pretty much all of the steps lol.

Thanks!

Edit: Using that link, I thought I could take the common factor of $1/\sqrt n$, but that just led to more confusion. And I also wasn't sure how I could implement $k/n$ into the equation.

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First, let's write the expression as a sum:

$$s_n=\sum\limits_{k = 1}^n {\frac{1}{{\sqrt {n + k} }}\frac{1}{{\sqrt n }}}$$

It is first stated that

$$\sum\limits_{k = 1}^n {f\left( {\frac{k}{n}} \right)\frac{1}{n} \to \int\limits_0^1 {f\left( x \right)dx} }$$

This means that the sum constructed on the left will tend to the value of the definite integral of $f$ over $[0,1]$. This is a result from Darboux/Riemann integration you might find in most textbooks. Assuming this result, we seek to use it to evaluate some sums. First, we need to write

$$s_n=\sum\limits_{k = 1}^n {\frac{1}{{\sqrt {n + k} }}\frac{1}{{\sqrt n }}}$$

as

$$\sum\limits_{k = 1}^n f\left( {\frac{k}{n}} \right) \frac{1}{n}$$

for some $f$. To find $f$, we must isolate the $1/n$ term, and see what is left. In this case:

\eqalign{ & {s_n} = \sum\limits_{k = 1}^n {\frac{1}{{\sqrt {n + k} }}\frac{1}{{\sqrt n }}} = \sum\limits_{k = 1}^n {\frac{1}{n}\frac{1}{{\sqrt {n + k} }}\frac{n}{{\sqrt n }}} = \cr & {s_n} = \sum\limits_{k = 1}^n {\frac{1}{n}\frac{{\sqrt n }}{{\sqrt {n + k} }}} = \sum\limits_{k = 1}^n {\frac{1}{n}\sqrt {\frac{n}{{n + k}}} } = \sum\limits_{k = 1}^n {\frac{1}{n}\frac{1}{{\sqrt {\frac{{n + k}}{n}} }}} = \sum\limits_{k = 1}^n {\frac{1}{n}\frac{1}{{\sqrt {1 + \frac{k}{n}} }}} \cr}

Can you take it from there?

Do not hover over the grey areas unless you want a solution. Try to think about it first.

So we can see that $f(x)=\frac{1}{\sqrt{1+x}}$. This means that $$\mathop {\lim }\limits_{n \to \infty } {s_n} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{1}{{\sqrt n }}\frac{1}{{\sqrt {n + k} }}} = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{1}{n}\frac{1}{{\sqrt {1 + \frac{k}{n}} }}} = \int\limits_0^1 {\frac{{dx}}{{\sqrt {1 + x} }}} = \frac{1}{2}\left( {1 - \sqrt 2 } \right)$$

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I'm confused on where you got the n/sqrt(n). I see how you have to isolate 1/n, but I'm not seeing how that somehow turns into n/sqrt(n) –  Josh Yard May 30 '12 at 0:40
$$\frac{1}{{\sqrt n }} = \frac{n}{n}\frac{1}{{\sqrt n }} = \frac{1}{n}\frac{n}{{\sqrt n }} = \frac{1}{n}\sqrt n$$ –  Pedro Tamaroff May 30 '12 at 0:42
I guess I understand how you got it, I'm just mostly confused on where it came from. In other words, what process did you do to get from the first to the second step underneath this line "we must isolate the 1/n term, and see what is left." Sorry If I seem stupid, which I probably am being right now. –  Josh Yard May 30 '12 at 0:54
@JoshYard I wanted a $1/n$ to appear. The simplest way was to actually multiply by $1/n$ and "cancel it" by multiplying by $n$. Then you move on with the solution by writing the rest as a function of $k/n$. –  Pedro Tamaroff May 30 '12 at 1:13

Hints:

1. You can factor one term in the denominators to get a $\frac{1}{n}$ factor in all the summands.
2. $[0,1]$ is not the only interval to integrate over.
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Yes, $[0,1]$ or change the function by $\frac{1}{\sqrt{x}}$ and integrate over $[1,2]$. It's the same. –  leo May 30 '12 at 0:36