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Problem :

Evaluate $$\lim_{n\to\infty} 2^{-n}\sum_{k_1 = 1}^2 \sum_{k_2 = 1}^2 \cdots \sum_{k_n = 1}^2 \frac{k_1 + k_2 + \cdots + k_n}{k_1 ^2 + k_2 ^2 + \cdots + k_n ^2}$$


This is one of problems in math contest but I can't find sol.

And I don't know where to start this, Can anyone give me hints?

If this problem related with probabilty or some disturibution, I want to know solution without using that, Thanks.

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Using only integral calculus: $$\begin{split} 2^{-n}\sum_{k_1= 1}^2 \cdots \sum_{k_n = 1}^2 \frac{k_1 + \cdots + k_n}{k_1 ^2 + \cdots + k_n ^2} &= 2^{-n}\sum_{k_1= 1}^2 \cdots \sum_{k_n = 1}^2\left (k_1 + \cdots + k_n\right)\int_0^{+\infty}e^{-(k_1 ^2 + + \cdots + k_n ^2)x}dx\\ &=2^{-n}\int_0^{+\infty}\sum_{k_1= 1}^2 \cdots \sum_{k_n = 1}^2\left (k_1 + \cdots + k_n\right)e^{-k_1^2x}\cdots e^{-k_n^2x}dx\\ &=2^{-n}\int_0^{+\infty}n\left(\sum_{k_1= 1}^2k_1e^{-k_1^2x}\sum_{k_2= 1}^2e^{-k_2^2x}\cdots\sum_{k_n= 1}^2e^{-k_n^2x}\right)dx\\ &=2^{-n}n\int_0^{+\infty}\left(e^{-x}+2e^{-4x}\right)\left(e^{-x}+e^{-4x}\right)^{n-1}dx\\ &=\int_0^{+\infty}\frac{e^{-\frac t n} + 2 e^{-\frac {4t} n}}{e^{-\frac t n}+e^{-\frac {4t}n}}e^{-t}\left(\frac{1+e^{-\frac {3t}n}}2\right)^n dt \,\,\,\text{ with }\,\,t=nx\\ \end{split}$$ By the dominated convergence theorem, $$\boxed{ 2^{-n}\sum_{k_1= 1}^2 \cdots \sum_{k_n = 1}^2 \frac{k_1 + \cdots + k_n}{k_1 ^2 + \cdots + k_n ^2} \longrightarrow \frac 3 2\int_0^{+\infty}e^{-t}e^{-\frac {3t} 2}dt=\frac 3 5}$$

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