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I'm asked to find $ \lim_{n\to\infty}a_n = \frac{1}{n^2} \sum_{k=1}^{n} k \cdot e^\frac{k^2}{n^2} $.

We can arrange the sum and we receive that our original limit is $ \frac{1}{n} \sum_{k=1}^{n} \frac{k}{n} \cdot e^\frac{k^2}{n^2} $.

Taking $ x_k = \frac{k}{n} $, that is the Riemann sum of the function $ xe^{x^2} $ on the interval $ [0,1] $. Since that function is continuous and Riemann integrable on $ [0,1] $, the original limit is equal to $ \int_{0}^{1} xe^{x^2} $. By that, the answer to this should be $ 0.5(e-1 )$. However, when looking up larger and larger values of $ n$, the limit tends to $ \frac{1}{2} $.

What is incorrect with my train of thought?

FNB
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    $\int_0^1 x e^{x^2}dx$ is correct, see https://math.stackexchange.com/q/1209823/42969 – Martin R Apr 21 '23 at 08:33
  • Oh, that is reassuring. Thank you! – FNB Apr 21 '23 at 08:34
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    How did you “look up larger and larger values of $n$”? I assume that there is an error in your calculation. – Martin R Apr 21 '23 at 08:36
  • I wrote the expression in Symbolab where $ n = 10^4, 10^6, 10^8 $... Are those not large enough values? Is the website not accurate enough? – FNB Apr 21 '23 at 08:46

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