Find $$\lim_\limits{n\to \infty}{b_n={1\over n^2}\sum_{i=1}^{n}ie^{\left(i^2\over n^2\right)}}$$. Trying to present it in the form of a Riemann sum, I have arrived at: $$\sum_{i=1}^{n}{1\over n}{ie^{\left(i^2\over n^2\right)}\over n}$$. Thus, $\Delta x={1\over n}$. However, I am not really sure how to actually pick $\Delta$, $x_i$ or $f(x)$. How should I do that? It is really confusing. I have to take $x_{i+1}-x_i$ into consideration, and $\Delta x$ as well. I would fully appreciate your help.
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Is it not $$\int xe^{x^2}dx$$ – Arpan Mar 28 '15 at 07:09
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$({i\over n})^2$ is in the exponent. – Meitar Abarbanel Mar 28 '15 at 07:11
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Not just $i$, although it looks that ways, as it is tiny. – Meitar Abarbanel Mar 28 '15 at 07:11
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I have edited it to look a little better. Is my previous comment correct? – Arpan Mar 28 '15 at 07:14
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You may write your sum as $$ \sum_{i=1}^{n}\frac{1}{n}\times\frac{i}{n}e^{\large \frac{i^2}{n^2}} $$ thus comparing with the following Riemann sum: $$ \sum_{i=1}^n (x_{i+1} - x_{i})\times f(x_i) $$ you get $$ x_i=\frac i n, \quad x_{i+1} - x_{i}=\frac 1 n, \quad f(x)=xe^{x^2}, $$ and, as $n \to +\infty$, you easily obtain $$ \lim_\limits{n\to \infty}{b_n}=\int_0^1xe^{x^2} dx=\frac 12\int_0^1e^{u} du=\frac{e-1}{2}. $$
Olivier Oloa
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