4

Find the value.

$\lim_{n \rightarrow \infty}$ $[(1 + 1/n^2)(1 + 2^2/n^2)^2 \cdots (1 + n^2/n^2)^n]^{1/n}$

I have tried. But I failed to establish any suitable proper form so that I can apply definition.

Manglu
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  • See also: https://math.stackexchange.com/questions/475786/how-to-compute-lim-n-rightarrow-infty-frac1n-left-2n12n2-cdots2nn and https://math.stackexchange.com/questions/1598508/i-need-help-to-advance-in-the-resolution-of-that-limit-lim-n-to-infty-s – lab bhattacharjee Jan 03 '18 at 10:54

1 Answers1

8

Note that we can take $\ln$ and write as:

$$\ln(L) = \lim_{n\to \infty}\frac{1}{n} \left( \sum_{i=1}^{n}i \ln(1+(\tfrac{i}{n})^2)\right) $$

We note that it cannot be written in the form of Riemann sum as there is an $n$ missing in the denominator.

As Jean recommends, had it been $1/n^2$ in the power, we could have written as:

$$\ln(L) = \lim_{n\to \infty}\frac{1}{n} \left( \sum_{i=1}^{n} \tfrac{i}{n} \ln(1+(\tfrac{i}{n})^2)\right)= \int_{0}^{1} x\ln(1+x^2)$$

Where we can compute the last integral as $$ \frac{(\ln(x^2+1))(x^2+1)-x^2}{2} |_{0}^{1} = \frac{\ln(4)-1}{2}$$

or that $L = \exp\left(\tfrac{\ln(4)-1}{2}\right)$. But our original sum diverges.

jonsno
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