Calculating $\lim\limits_{n\to\infty}\ln\sqrt[n]{(1+\frac{1}{n^2})(1+\frac{4}{n^2})\cdots(1+\frac{(n-1)^2}{n^2})(1+\frac{n^2}{n^2})}$

Question

Find$$\lim_{n\to\infty}\ln\sqrt[n]{\left(1+\frac1{n^2}\right)\left(1+\frac4{n^2}\right)\cdots\left(1+\frac{(n-1)^2}{n^2}\right)\left(1+\frac{n^2}{n^2}\right)}$$ and $$\lim_{n\to\infty}\ln\sqrt[n]{\left(1+\frac1n\right)\left(1+\frac2n\right)\cdots\left(1+\frac{n-1}n\right)\left(1+\frac nn\right)}.$$

I know I have to first use logarithm and with that in both I got $$\frac{1}{n}\lim_{n \rightarrow \infty}ln(1+\frac{1}{n^2})+ln(1+\frac{4}{n^2})+...+ln(1+\frac{(n-1)^2}{n^2})+ln(1+\frac{n^2}{n^2})$$ and $$\frac{1}{n}\lim_{n \rightarrow \infty}ln(1+\frac{1}{n})+ln(1+\frac{2}{n})+...+ln(1+\frac{(n-1)}{n})+ln(1+\frac{n}{n})$$

Any help?

Answer 1

The first one:

$$\lim_{n\to \infty}\dfrac 1 n \sum_{i=1}^n \ln\left(1+\left(\dfrac i n\right)^2\right)=\int_0^1 \ln(1+x^2)dx$$

The second one:

$$\lim_{n\to \infty}\dfrac 1 n \sum_{i=1}^n \ln\left( 1+\dfrac i n \right) =\int_0^1 \ln(1+x)dx$$

Answer 2

Notice that you may write your limit as follows:

$$ \lim_{n \to \infty} \frac{1}{n} \sum_{i=1}^n \ln \left( 1 + \dfrac{i}{n} \right) = \int\limits_0^1 \ln (1+x) dx = x \ln(1+x) \bigg|_0^1 - \int\limits_0^1 \dfrac{xdx}{x+1} = \ln 2 - \int\limits_0^1 dx + \int\limits_0^1 \dfrac{dx}{x+1} = \ln 2 - 1 + \ln 2 = \boxed{ \ln 4 -1 }$$

As for the first one:

$$\lim_{n\to \infty}\dfrac 1 n \sum_{i=1}^n \ln\left(1+\left(\dfrac i n\right)^2\right)=\int\limits_0^1 \ln(1+x^2)dx = x \ln (1+x^2) \bigg|_0^1 - \int\limits_0^1 \dfrac{2 x^2 }{1+x^2}dx = \ln 2 -2 \int\limits_0^1 dx + 2 \int\limits_0^1 \dfrac{dx}{1+x^2} = \ln 2 - 2 + 2 \arctan(1) - 2 \arctan(0) = \boxed{ \ln 2 -2 + \dfrac{ \pi }{2} }$$

Answer 3

You want $$\lim_{n\to+\infty}\frac 1n\sum_{k=1}^n\ln(1+(\frac kn)^2)$$

this is a Riemann sum of the fonction $$x\mapsto \ln(1+x^2)$$ over $[0,1]$.

$$\int_0^1\ln(1+x^2)dx=\Bigl[x\ln(1+x^2)\Bigr]_0^1-\int_0^1x\frac{2x}{1+x^2}dx$$

$$=\ln(2)-2\int_0^1\frac{x^2+1-1}{x^2+1}dx$$ $$=\ln(2)-2+\frac{\pi}{2}$$

For the second, it is $$\int_0^1\ln(1+x)dx=$$ $$\Bigl[(x+1)\ln(x+1)-x\Bigr]_0^1=$$

$$2\ln(2)-1$$

Answer 4

Since you already received good answers for the limit itself, just for the fun of it, I tried to compute the partial terms

$$S_n=\frac 1n \log\left(\prod_{k=1}^n 1+\frac {k^2}{n^2}\right)=\frac 1n \log\left(n^{-2 n} (1-i n)_n (1+i n)_n\right)$$ where appear Pochhammer symbols. Expanding the logarithm and using asymptotics of Pochhammer symbols, we easily obtain $$S_n=\left(\frac{\pi }{2}+\log (2)-2\right)+\frac{\log (2)}{2 n}+\frac{1}{12 n^2}+\frac{1}{720 n^4}+O\left(\frac{1}{n^6}\right)$$ which is almost exact even for small values of $n$ $$\left( \begin{array}{ccc} n & \text{approximation} & \text{exact} \\ 5 & 0.336593780966392 & 0.336593768083114 \\ 10 & 0.299434338605061 & 0.299434338405910 \\ 15 & 0.287418811178719 & 0.287418811161272 \\ 20 & 0.281480528882729 & 0.281480528879626 \\ 25 & 0.277939787854930 & 0.277939787854117 \end{array} \right)$$