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Is it possible to do such a sum?

$$\sum_{i=1}^n \sqrt (i^2+\frac{8i}{n}+\frac{16+n^2}{n^2})$$

I want to reach to a function with only the n as variable. I believe that it is possible somehow by squaring the summation and squaring the other side of the equation.

For more specification, this summation is a part of:

$$C = \lim_{n \to ∞} [\frac{2}{n}\sum_{i=1}^n \sqrt (i^2+\frac{8i}{n}+\frac{16}{n^2}+1)]$$

Mac Spour
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1 Answers1

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Computing the sum is not needed to determine the value of $C$. To do the latter, note that $$i^2+\frac{8i}n+\frac{16}{n^2}+1\geqslant i^2$$ hence $$\frac2n\sum_{i=1}^n\sqrt{ i^2+\frac{8i}n+\frac{16+n^2}{n^2}}\geqslant\frac2n\sum_{i=1}^ni=n+1$$ hence $$C=+\infty$$ Edit: On the other hand, $$\frac2n\sum_{i=1}^n\sqrt{ 1+\frac{4i^2}{n^2}}$$ is the $n$th Riemann sum of the function $f(x)=\sqrt{1+x^2}$ on $[0,2]$ hence its limit when $n\to\infty$ is direct but there is no reason to expect a closed form for these sums for every finite $n$.

Did
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  • Woah, i think there is a misconception here, I thought that I was going to find the length of an arch. That is strange to be honest. $$C = \lim_{n \to ∞} [\Delta x \sum_{i=1}^n \sqrt (1+f'(xi)^2 ]$$ this is the formula i concluded from reimann summs to find the arch length. and the original formula i want to calculate was $$f(x) = x^2 $$ and the domain was [0,2] – Mac Spour Mar 05 '18 at 16:44
  • Is there a mistake in the formula in your question? Maybe $i^2$ should read $\left(\frac in\right)^2$? – Did Mar 05 '18 at 16:47
  • Well, to find $f'(xi)$ i found $xi$ itself; $$xi= xo + \Delta xi$$ . The $\Delta xi$ is calculated like this $$ \Delta xi = \frac{b-a}{n} = \frac{2-0}{n}$$ . Now regarding that $$f'(x) = 2x$$ , $f'(xi)$ should be $ \frac{2}{n} i$ , squaring it makes it $ \frac {(4+ni)^2}{n^2}$ oui I just realized that squaring it is not $ \frac {(4+ni)^2}{n^2}$ but $ \frac {4}{n^2} i^2$ – Mac Spour Mar 05 '18 at 16:56
  • Well just nevermind that, I really apologize, that's some gore over there... – Mac Spour Mar 05 '18 at 17:03
  • See edit. The idea to compute the exact value of every Riemann sum is odd anyway, and rather useless if you ask me. – Did Mar 05 '18 at 17:05
  • Thanks a lot, I really appreciate your help. I just took the Riemann sum and integration last week and I thought that it could be possible to calculate the length using the same exact method. Well That is reasonable as you said that there is no need to limit the $n$ towards infinity, just a really large number and a calculator shall do. – Mac Spour Mar 05 '18 at 17:10
  • My point was rather that there is little hope to compute the value of $$I=\int_0^2\sqrt{1+x^2}dx$$ by computing the values of each of the Riemann sums. In general, the latter is a more difficult task than the former. And as a matter of fact, standard change of variable techniques yield, modulo some mistake from my part, the value $$I=\sqrt 5+\tfrac12\ln(2+\sqrt 5)$$ – Did Mar 05 '18 at 17:17
  • Is there a better way to compute the length of a function's graph? – Mac Spour Mar 05 '18 at 17:28
  • Than to compute the corresponding integral? It seems not. – Did Mar 05 '18 at 17:28