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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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    this is just a riemann sum integral 0 to 1 of 1/sqrt.1+x2012-12-28
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    @Rustyn: $$\sum_{k=1}^n\frac{1}{\sqrt{n^2}}=\sum_{k=1}^n\frac{1}{n}=\underbrace{\frac{1}{n}+\cdots+\frac{1}{n}}_{n\text{ times}}=1$$2012-12-28
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    @K.Ghosh This comment worth for an answer. Why don't you post it as an answer.2012-12-28
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    Something is wrong here. How are you summing from $n$ to $n$?2012-12-28
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    @ZevChonoles Thx. I was confused by indexing variables.2012-12-28

1 Answers 1

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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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    And when you get the result, you may indeed verify that it satisfies the inequalities you derived.2012-12-28
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    can you plz. solve it ...2012-12-28
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    @ram : you just need to find $f(x)$ and integral limits $a,b$. Then evaluate the integral. One more hint that $f(x)=\frac{1}{\sqrt{x}}$. Can you find the result now?2012-12-28
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    Thanks Sir !!! I have the answer $2\sqrt {2} - 2$.2012-12-28
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    @ram That's right. You got it.2012-12-28
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    @Mathlover shouldnot $f(x)$ be $\frac{1}{\sqrt\(1+x\)}$2012-12-28
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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