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$$\lim_{n\rightarrow\infty}\sum_{i=1}^n\left(\sqrt{8+\frac {4i}n}\right)\frac 4n$$

I know this is the Riemann sum for certain integral, and then the limit is just the integral, but is there any way to solve this without using integrals?

This question is in a guide of problems for students who do not know integrals yet.

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    $\begingroup$ Woah boy... now I really want to start a movement forbidding $i$ from being used as a summation index.... ;) I was wondering for the longest time how the imaginary unit into the integral... $\endgroup$
    – apnorton
    Commented Jan 11, 2013 at 2:40

2 Answers 2

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$$4\left[\frac{1}{n}\sum_{k=1}^n\left(\sqrt{8+\frac{4k}{n}}\right)\right]\xrightarrow[n\to\infty]{}4\int\limits_0^1\sqrt{8+4x}\,dx$$

Check the above with the partition $\,\left\{0<\frac{1}{n}<\frac{2}{n}<\ldots <\frac{n}{n}=1\right\}\,$

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  • $\begingroup$ I believe what you wrote above is off by a a factor of $4$. Also it seems that the OP is asking for a solution that does not use the integral. $\endgroup$
    – Eric Naslund
    Commented Jan 11, 2013 at 3:13
  • $\begingroup$ @EricNaslund, thanks. About your first comment I think the answer is correct as the whole parentheses in the sum is multiplied by $\,4/n\,$ , which isn't affected by the sum's index. About the second comment: right, and I'm going to delete my answer in a little while. $\endgroup$
    – DonAntonio
    Commented Jan 11, 2013 at 3:15
  • $\begingroup$ You are correct, sorry. I had the the integral was $\int_0^4 \sqrt(8+x)dx$, but I didn't look at your integrand properly. $\endgroup$
    – Eric Naslund
    Commented Jan 11, 2013 at 3:17
  • $\begingroup$ Please leave your answer up... that way, it's easier to check each others' work when they post a summation... $\endgroup$
    – apnorton
    Commented Jan 11, 2013 at 3:42
  • $\begingroup$ Ok, @anorton: done. $\endgroup$
    – DonAntonio
    Commented Jan 11, 2013 at 3:43
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Let $$S_n=\sqrt1+\sqrt2+\cdots+\sqrt n$$ By Stolz-Cesàro theorem, we have $$\lim_{n\to\infty}\frac{\sqrt1+\cdots+\sqrt n}{n\sqrt n}=\lim_{n\to\infty}\frac{\sqrt n}{n\sqrt n-(n-1)\sqrt{n-1}}=\lim_{n\to\infty}\frac{n^2\left(1+\left(1-\frac1n\right)\sqrt{1-\frac1n}\right)}{n^3-(n-1)^3}=\frac23$$ Therefore $$S_n\sim \frac23n\sqrt n$$ Notice that $$\frac1n\sum_{k=1}^n\sqrt{8+\frac{4k}n}=\frac2{n\sqrt n}(S_{3n}-S_{2n})$$ Thus $$\lim_{n\to\infty}\frac4n\sum_{k=1}^n\sqrt{8+\frac{4k}n}=\frac{16}3(3\sqrt3-2\sqrt2)$$

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  • $\begingroup$ This assumes that one knows that $n\sqrt{n}-(n-1)\sqrt{n-1}\sim\frac32\sqrt{n}$. Derivative? $\endgroup$
    – Did
    Commented Jan 11, 2013 at 8:31
  • $\begingroup$ @did If $(1+x)^\alpha-1\sim\alpha x$ is allowed, then it's not hard to show that. Well, suppose that we only have tools of elementary mathematics and some basic properties of a limit of sequence, it's a bit more tricky: $n\sqrt n-(n-1)\sqrt{n-1}=(n^3-(n-1)^3)/(n\sqrt n+(n-1)\sqrt{n-1})=(3n^2-3n+1)/(n\sqrt n+(n-1)\sqrt{n-1})$. Now it's not too difficult to obtain the result. $\endgroup$
    – Yai0Phah
    Commented Jan 11, 2013 at 8:38
  • $\begingroup$ Right. I was not asking for a way to prove this result, but suggesting that, to make the answer coherent, one needed a way to prove it according to the OP's rules (hence I suggest to include in your answer the argument you just explained). $\endgroup$
    – Did
    Commented Jan 11, 2013 at 8:43

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