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Determine the following limit by interpreting it as a definite integral: $$\displaystyle\lim_{n\to\infty}\frac{1}{n} \left(\frac{1}{(1+1/n)^2}+\frac{1}{(1+2/n)^2}+\dots +\frac{1}{(1+n/n)^2}\right)$$ So, the answer is $$\int\limits_{1}^{2} \frac{1}{x^2}$$ and I can see that by inspection, but what steps are involved between the two? The integral is equal to $$\displaystyle\lim_{n\to\infty}\sum_{i=1}^{n}\frac{1}{n} \left(\frac{1}{(1+i/n)^2}\right)$$ But how am I supposed to 'guess' that $(1+i/n)=x$ where $x\in[1,2]$? Does $\displaystyle \lim_{i\to n}\frac{i}{n} =$ the range, and +1 the lower limit?

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Consider function $f(x)=\frac{1}{x^2}$ and partition $y_0 < y_1< \cdots <y_n$ where $y_i=1+\frac{i}{n}$. Riemann sum for this partition is:

$\sum_{i=0}^{n}\frac{1}{n} \frac{1}{(1+\frac{i}{n})^2}=\frac{1}{n}\sum_{i=0}^{n}\frac{1}{(1+\frac{i}{n})^2}$

Now we know that $f$ is continuous on $[1,2]$, so it's Riemann integrable. So when $\max\{ y_{i+1}-y_i : 0 \leq i<n\} \to 0$ the Riemann's sums converge to integral $\int_{1}^{2} f(x) dx$, but when $n \to \infty$ we have $\max\{ y_{i+1}-y_i : 0 \leq i<n\} \to 0$, so (for example by http://en.wikipedia.org/wiki/Riemann_integral):

$\lim_{n\ \to \infty} \frac{1}{n}\sum_{i=0}^{n}\frac{1}{(1+\frac{i}{n})^2}=\int_{1}^{2} f(x) dx$

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