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How can I find the limit $\lim_{n\to\infty}\displaystyle\sum_{k=1}^n\frac{1}{\sqrt {n^2 +kn}} \quad?$

I have tried to solve it using squeeze theorem: $\sum_{k=1}^n\frac{1}{\sqrt {n^2 +kn}} > \displaystyle\sum_{k=1}^n\frac{1}{\sqrt {n^2 +n^2}} = \displaystyle\sum_{k=1}^n\frac{1}{\sqrt {2n^2}}=\frac{1}{\sqrt {2}} $ and $\sum_{k=1}^n\frac{1}{\sqrt {n^2 +kn}} <\displaystyle\sum_{k=1}^n\frac{1}{\sqrt {n^2}} = 1.$

But I could not find the sequences with the same limits.

Please help - how to solve this?

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    @ZevChonoles Thx. I was confused by indexing variables.2012-12-28

2 Answers 2

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HINT:

$\int _a^b {f(x) dx}=\lim_{n\to\infty} \frac{b-a}{n}\sum \limits_{k=1}^n f\left(a+\frac{k(b-a)}{n}\right) \tag 1$

$\lim_{n\to\infty}\displaystyle\sum_{k=1}^n\frac{1}{\sqrt {n^2 +kn}} =\lim_{n\to\infty} \frac{1}{n}\displaystyle\sum_{k=1}^n\frac{1}{\sqrt {1 +k\frac{1}{n}}} \tag 2$

Can you proceed after that?

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    @jim it can be selected such. if you select $f(x)=\frac{1}{\sqrt{x+1}}$ you must use $a=0$ and $b=1$. In my case, $a=1$ , $b=2$2012-12-28