Finding the limit of the given series

Question

What is the best way to find the following limit:

$$\lim_{n \to \infty} \frac{((n+1)(n+2)\cdots(n+n))^{\frac1n}}{n} $$

And what is the final value?

I tried finding patterns of $\frac i n$ in the limit, but haven’t reached any conclusion.

Answer 1

$$\lim_{n \to \infty} \frac{((n+1)(n+2)\cdots(n+n))^{\frac1n}}{n}=\lim_{n\rightarrow+\infty}\prod_{k=1}^n\left(1+\frac{k}{n}\right)^{\frac{1}{n}}=$$ $$=e^{\frac{1}{n}\lim\limits_{n\rightarrow+\infty}\sum\limits_{k=1}^n\ln\left(1+\frac{k}{n}\right)}=e^{\int\limits_0^1\ln(1+x)dx}=e^{((1+x)\ln(1+x)-x)|_0^1}=e^{2\ln2-1}=\frac{4}{e}.$$

Answer 2

$$\ln L=\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^{\infty} \ln (1+\frac{k}{n})= \int_{0}^{1} \ln (1+x) dx=(1+x)\log(1+x)-(1+x)|_{0}^{1}= 2\ln 2-1$$ $$\implies L=\frac{4}{e}$$

Answer 3

Using this property, it is enough to find

• $\lim_{n\to \infty}\frac{a_{n+1}}{a_n}$

with $a_n = \frac{(2n)!}{n!n^n}$.

$$\frac{a_{n+1}}{a_n }= \frac{2(n+1)(2n+1)}{(n+1)^2\left(1+\frac 1n\right)^n}\stackrel{n\to\infty}{\longrightarrow}\frac 4e$$

Answer 4

More than likely, too complex (but I had fun !)

If you are familiar with Pochhammer symbols and the gamma function $$P_n=\prod_{i=1}^n (n+i)=(n+1)_n=\frac{2^{2 n} }{\sqrt{\pi }}\Gamma \left(n+\frac{1}{2}\right)$$ Take logarithms, use the asymptotics (Stirling formula) $$\log (\Gamma (p))=p (\log (p)-1)+\frac{1}{2} \left(\log (2 \pi )-\log(p)\right)+\frac{1}{12 p}+O\left(\frac{1}{p^3}\right) $$ Apply it and continue with Taylor expansion to get $$\log(P_n)=n (\log (n)-1+2 \log (2))+\frac{\log (2)}{2}-\frac{1}{24 n}+O\left(\frac{1}{n^3}\right)$$ $$\frac 1n \log(P_n)=(\log (n)-1+2 \log (2))+\frac{\log (2)}{2 n}-\frac{1}{24 n^2}+O\left(\frac{1}{n^4}\right)$$ $$P_n^ {\frac 1n}=e^{\frac 1n \log(P_n)}=\frac{4 n}{e}+\frac{2 \log (2)}{e}+\frac{3 \log ^2(2)-1}{6 e n}+O\left(\frac{1}{n^2}\right)$$ $$\frac {P_n^ {\frac 1n}}n=\frac{4 }{e}+\frac{2 \log (2)}{e n}+\frac{3 \log ^2(2)-1}{6 e n^2}+O\left(\frac{1}{n^3}\right)$$ which shows the limit and how it is approached.

Moreover, this gives a shortcut method for the evaluation of the expression. For example $P_5=1.574513$ while the truncated expansion would give $1.574598$.