$$\lim_{n\to \infty} \sum _{i=1}^{n} (\sqrt{n}+\frac{i}{\sqrt{n}})^{-2}$$

I am asked to find the limit of this sum and my prof wants me to interpret it as a definite integral.

Treating it as a Riemann sum I would need to find $\Delta{x}$ and $ x_i^\ast$ and find the function, but I don't really know how to approach the problem. Thank you for the help.


Hint: Write

$$\lim_{n\to \infty} \sum _{i=1}^{n} \left(\sqrt{n}+\frac{i}{\sqrt{n}}\right)^{-2} = \lim_{n\to\infty} \frac 1n \sum_{i=1}^n \frac{1}{(1 + i/n)^2}$$

See what to do now?

  • $\begingroup$ Is the function $\frac{1}{x^2}$ and the limits of integration from $1$ to $2$? $\endgroup$ – Daniel Waleniak Dec 4 '15 at 23:35
  • $\begingroup$ That works. Another (and perhaps more natural) choice is $1/(1+x)^2$ from $0$ to $1$. $\endgroup$ – Simon S Dec 4 '15 at 23:36
  • 2
    $\begingroup$ FYI, the thinking to get there was "I want a $1/n$ term out the front. What happens when I force that? Ah, it works nicely" $\endgroup$ – Simon S Dec 4 '15 at 23:38

Notice, $$\lim_{n\to \infty}\sum_{i=1}^n\left(\sqrt n+\frac{i}{\sqrt n}\right)^{-2}$$ $$=\lim_{n\to \infty}\sum_{i=1}^n\frac{1}{n}\left(1+\frac{i}{n}\right)^{-2}$$ let $\frac{i}{n}=x\implies \frac{1}{n}=dx$ & using integration with proper limits, $$=\int_{0}^1(1+x)^{-2}\ dx$$ $$=\left(\frac{-1}{1+x}\right)_0^1$$ $$=-\frac{1}{2}+1=\color{red}{\frac{1}{2}}$$

  • 1
    $\begingroup$ Got the same result using Simon's hint. Thanks Harish. $\endgroup$ – Daniel Waleniak Dec 4 '15 at 23:51

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