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Find the limit as $n$ approaches infinity: $\lim_{n\to \infty}$$\sum_{i=1}^n (\frac{1}{\sqrt{n}+\frac{i}{\sqrt{n}}})^2$

I can't seem to be able to figure out how to solve these with the i in the denominator, using the wolfram calculator I got an advanced answer with some gamma notation in it that makes no sense to me. How do I solve it without the gamma notation.

I've gotten it down to limit as $n$ approaches infinity: $\lim_{n\to \infty}$$\sum_{i=1}^n (\frac{n}{n^2+2in+i^2})$

I am only in calculus 1 and am looking for a way to seperate the function to where i can apply the sigma notation to each part of the function using the formulas for the sums of powers

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Note that we can write

$$\begin{align} \sum_{k=1}^{n} \left( \frac{1}{\sqrt n +k/\sqrt n} \right)^2&=\sum_{k=1}^{n} \left( \frac{\sqrt n}{n +k} \right)^2\\\\ &=\sum_{k=1}^n\frac{n}{(n+k)^2}\\\\ &=\sum_{k=n+1}^{2n}\frac{n}{k^2} \end{align}$$

Then, note that since $1/k$ monotonically decreases, we have the bounds (SEE THIS)

$$\int_{n+1}^{2n+1} \frac{n}{x^2}\,dx\le \sum_{k=n+1}^{2n}\frac{n}{k^2}\le \frac1{(n+1)^2}+\int_{n+1}^{2n}\frac{n}{x^2}\,dx$$

Finish by applying the squeeze theorem.

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One may recognize a Riemann sum by writing, as $n \to \infty$, $$ \sum_{k=1}^n\left(\frac{1}{\sqrt{n}+\frac{k}{\sqrt{n}}}\right)^2=\frac1n \cdot\sum_{k=1}^n\frac{1}{\left(1+\frac{k}{n}\right)^2} \to \int_0^1\frac{dx}{\left(1+x\right)^2}. $$

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