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$\displaystyle f(x)=\lim \limits_{n\rightarrow \infty} \left(\frac{n^n(x+n)(x+n/2)...(x+n/n)}{n!(x^2+n^2)(x^2+n^2/4)...(x^2+n^2/n^2)}\right)^{x/n}$, find $f(1/3)$

$f(x)=\displaystyle \left({\frac{\prod \limits_{r=1}^n (1+xr/n)}{\prod \limits_{r=1}^n (1+(xr/n)^2)}}\right)^{x/n}=e^{x/n \ln{\frac{\prod \limits_{r=1}^n (1+xr/n)}{\prod \limits_{r=1}^n (1+(xr/n)^2)}}}$

We can differentiate but in the exam we're expected to do this within $10$ minutes, there must be another shorter method to simplify this

Tatai
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  • You may use definite integrals as limit of a sum after splitting $\log(a/b)$ as $log(a)-log(b)$. – Ilovemath Sep 04 '21 at 06:59
  • This has been asked and answered before: https://math.stackexchange.com/q/2312997/42969, https://math.stackexchange.com/q/2223400/42969. – Martin R Sep 04 '21 at 07:03
  • Also: https://math.stackexchange.com/q/2274025/42969 (with links to more identical questions). – Martin R Sep 04 '21 at 07:06

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