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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.

Any advice?

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.

2 Answers 2

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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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    @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 $k/n$.2012-05-30
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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.2012-05-30