PhD admission product $\lim_{n\to 0}\left(\frac21\left(\frac32\right)^2\left(\frac43\right)^3\cdots\left(\frac{n+1}{n}\right)^n\right)^{1/n}$

Question

Hello there I saw this problem (#3) here: http://www.sau.int/admission/2018/samplepapers/PAM.pdf

$$L=\lim_{n\to 0} \left( \frac{2}{1}\left(\frac{3}{2}\right)^2\left(\frac{4}{3}\right)^3...\left(\frac{n+1}{n}\right)^n\right)^\frac{1}{n}$$ The choices for the answer are $e$, $\pi$, $\frac{1}{e}$, $\frac{1}{\pi}$.

If we take the logartihm on both sides we get: $$\ln L=\frac{1}{n}\sum_{k=1}^{n} k\ln\frac{k+1}{k}$$ thus by telescoping $$\ln L= \ln\left(\frac{n+1}{(n!)^{\frac{1}{n}}}\right)$$ and now using wolfram I get the answer to be $L=e^\gamma$, which is not one of the choices.

Where did I go wrong? And could you share the correct solution?

Answer 1

$$\dfrac{n+1}{(n!)^{\frac{1}{n}}} \approx \dfrac{n+1}{\sqrt{2 \pi n}^{1/n}\dfrac n e} \to e$$

Answer 2

Hint

after simplification,

$$\ln (L)=\lim_{n\to+\infty}\Bigl (\ln(n+1)-\frac {1}{n}\ln (n!)\Bigr)= $$

Answer 3

Of the sequence $a_n=(1+\frac1n)^n$ which we know converges to $e$ the task asks you to consider the geometric mean sequence $$ b_n=\sqrt[n]{a_1a_2\cdots a_n}. $$ The logarithm of that has the form $$ \ln b_n=\frac{\ln a_1+\ln a_2+\dots+\ln a_n}n $$ which by the theorem about the Cesaro mean or by the more general theorem of Cesaro-Stolz converges to the same limit as the quotient sequence $\frac{\ln a_n}1$, if that exists. As we know that this limit exists and is $1$, the sequence $\ln b_n$ has the same limit and thus $b_n$ converges to $e$.