Calculate the limit of the following sequence: $\lim\limits_{n\to \infty} a_n = \frac{1}{n^2}\sum\limits_{i=1}^ni\cdot e^\frac{i^2}{n^2}$
Any help or hint would be appreciated.
Asked
Active
Viewed 108 times
2
-
5Hint: move one of the factors of $\frac1n$ into the summation and that sure does start to look an awful lot like a Riemann sum... – Steven Stadnicki Apr 11 '18 at 16:36
-
Can it be solved without using Riemann sums? – Galush Balush Apr 11 '18 at 16:43
-
3Given the quadratic term in the exponential, I'm hard-pressed to imagine any way that wouldn't just basically go through the integral. Why do you want to solve it without Riemann sums? Where did you come across this problem in the first place? – Steven Stadnicki Apr 11 '18 at 17:02
-
Try to use this property https://math.stackexchange.com/questions/2118515/if-f-is-riemann-integrable-on-0-1-then-lim-frac1n-sum-fk-n-int/ – rtybase Apr 11 '18 at 17:07
-
Does this answer your question? Find $\lim_\limits{n\to \infty}{b_n={1\over n^2}\sum_{i=1}^{n}ie^{i^2\over n^2}}$. – Anne Bauval Apr 21 '23 at 08:34
1 Answers
4
Write $$ a_n := \frac{1}{n^2} \sum \limits_{k = 1}^n k \cdot e^{\frac{k^2}{n^2}} = \frac{1}{n} \sum \limits_{k = 1}^n \frac{k}{n} \cdot e^{(\frac{k}{n})^2} $$ This is a Riemann sum for the (everywhere defined and continuous!) function $f(x) := x \cdot e^{x^2}$ over the interval $[0,1]$, hence the limit you're looking for exists and is given by $$ \lim \limits_{n \to \infty} a_n = \int \limits_0^1 x \cdot e^{x^2} dx $$ Elementary integration (e.g. using the substitution $y = x^2$) yields $$ \lim \limits_{n \to \infty} a_n = \int \limits_0^1 x \cdot e^{x^2} dx = \left[ \frac{1}{2} e^{x^2} \right]_0^1 = \frac{1}{2}(e-1) $$
ComplexF
- 725