How to find limit of the sequence $\sum\limits_{k=1}^n\frac{1}{\sqrt {n^2 +kn}}$?

Question

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?

@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 06:39
@K.Ghosh This comment worth for an answer. Why don't you post it as an answer. — Norbert, Dec 28 '12 at 06:39
Something is wrong here. How are you summing from $n$ to $n$? — Pricklebush Tickletush, Dec 28 '12 at 06:37

score 7 · Accepted Answer · edited Jan 28 '13 at 09:40

7

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?

edited Jan 28 '13 at 09:40

Norbert

56,803

answered Dec 28 '12 at 06:42

Mathlover

10,058

And when you get the result, you may indeed verify that it satisfies the inequalities you derived. – Ron Gordon Dec 28 '12 at 06:46
can you plz. solve it ... – ram Dec 28 '12 at 06:50
@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 06:55
Thanks Sir !!! I have the answer $2\sqrt {2} - 2$. – ram Dec 28 '12 at 07:01
@ram That's right. You got it. – Mathlover Dec 28 '12 at 07:02
@Mathlover shouldnot $f(x)$ be $\frac{1}{\sqrt(1+x)}$ – jim Dec 28 '12 at 07:04
@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, $a=1$ , $b=2$ – Mathlover Dec 28 '12 at 07:07
@Mathlover may you please derive how for a=0 we get the f(x) =1/√x form ? And for a=1 we get f(x) =1/√x+1 ? I always gets confused with this – ProblemDestroyer Apr 22 '22 at 21:31

score 1 · Answer 2 · answered Sep 27 '19 at 03:06

In comments to this recent question, this problem was mentioned and, being concerned by asymptotics, I tried to use a similar approach. $$S_n=\sum\limits_{k=1}^n\frac{1}{\sqrt {n^2 +kn}}=\frac{\zeta \left(\frac{1}{2},n+1\right)-\zeta \left(\frac{1}{2},2 n+1\right)}{\sqrt{n}}=\frac{H_{2 n}^{\left(\frac{1}{2}\right)}-H_n^{\left(\frac{1}{2}\right)}}{\sqrt{n}}$$ where appear the Hurwitz zeta function and generalized harmonic numbers. Using the asymptotics $$H_{p}^{\left(\frac{1}{2}\right)}=2 \sqrt{p}+\zeta \left(\frac{1}{2}\right)+\frac 1 {2\sqrt p}-\frac 1{24p\sqrt p}+O\left(\frac{1}{p^{7/2}}\right)$$ Apply it twice and continue with Taylor series to get $$S_n=(2 \sqrt{2}-2) \left(1-\frac{1}{4 \sqrt{2} n}+\frac{2+3 \sqrt{2}}{192 n^2}+O\left(\frac{1}{n^4}\right) \right)$$ For the fun of it, let us compute using $n=10^k$ to get $$\left( \begin{array}{ccc} k & \text{approximation} & \text{exact} \\ 0 & 0.70891579073141078638 & 0.70710678118654752440 \\ 1 & 0.81405181655943694309 & 0.81405158010700953614 \\ 2 & 0.82696535217966202992 & 0.82696535215592309803 \\ 3 & 0.82828070507205876331 & 0.82828070507205638932 \\ 4 & 0.82841248035460217890 & 0.82841248035460217866 \\ 5 & 0.82842566028278955788 & 0.82842566028278955788 \\ 6 & 0.82842697829960762615 & 0.82842697829960762615 \end{array} \right)$$

How to find limit of the sequence $\sum\limits_{k=1}^n\frac{1}{\sqrt {n^2 +kn}}$?

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