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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+x – Koushik Dec 28 '12 at 6:36
Something is wrong here. How are you summing from $n$ to $n$? – Pricklebush Tickletush Dec 28 '12 at 6:37
@Rustyn: $$\sum_{k=1}^n\frac{1}{\sqrt{n^2}}=\sum_{k=1}^n\frac{1}{n}=\underbrace{\frac{1}{‌​n}+\cdots+\frac{1}{n}}_{n\text{ times}}=1$$ – Zev Chonoles Dec 28 '12 at 6:39
@K.Ghosh This comment worth for an answer. Why don't you post it as an answer. – Norbert Dec 28 '12 at 6:39
@ZevChonoles Thx. I was confused by indexing variables. – Rustyn Dec 28 '12 at 6:42
up vote 3 down vote accepted


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

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

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. – Ron Gordon Dec 28 '12 at 6:46
can you plz. solve it ... – ram Dec 28 '12 at 6:50
@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? – Mathlover Dec 28 '12 at 6:55
Thanks Sir !!! I have the answer $2\sqrt {2} - 2$. – ram Dec 28 '12 at 7:01
@ram That's right. You got it. – Mathlover Dec 28 '12 at 7:02

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