$$ \lim_{x \rightarrow \infty} x\left(\frac{1}{x^2}+\frac{1}{(x+1)^2}+...+\frac{1}{(2x-1)^2}\right)$$
My answer is 7/24, but the correct answer provided by the book is 1/2.
Could anyone help me? Thanks for any insights.
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2How did you get 7/24? Why not show your work here? – Matthew Conroy Nov 14 '13 at 18:59
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$$\lim_{n\to\infty} n\left(\frac1{(n+0)^2}+\frac1{(n+1)^2}+\frac1{(n+n-1)^2}\right)$$
$$=\lim_{n\to\infty} n\left(\frac1{(n+1)^2}+\frac1{(n+1)^2}+\frac1{(n+n)^2}\right)-\lim_{n\to\infty} n\left(\frac1{(2n)^2}-\frac1{n^2}\right)$$
$$=\lim_{n\to\infty}\frac1n\sum_{1\le r\le n}\frac{n^2}{(n+r)^2}-0$$
$$=\lim_{n\to\infty}\frac1n\sum_{1\le r\le n}\frac1{\left(1+\frac{r}n\right)^2}$$
$$=\int_0^1\frac1{(1+x)^2}dx$$
$$\text{as }\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$
lab bhattacharjee
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@user2967915, my pleasure. See also , http://math.stackexchange.com/questions/465075/find-lim-limits-n-to-infty-frac1n-sum-limits2n-r-1-fracr-sq and http://math.stackexchange.com/questions/469885/the-limit-of-a-sum – lab bhattacharjee Nov 14 '13 at 19:13
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