Compute: $\lim_{n\to\infty} \left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)^{\frac{1}{2}}\cdots\left(1+\frac{n}{n}\right)^{\frac{1}{n}}$

Question

Compute the limit:

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

Answer 1

Note at the onset that $1+\frac{k}n\leqslant\mathrm e^{k/n}$ for every $k$ hence the $n$th product $P_n$ is such that $P_n\leqslant\mathrm e$, in particular, the sequence $(P_n)_{n\geqslant1}$ is bounded.

To show that $(P_n)_{n\geqslant1}$ actually converges and to identify its limit, note that, for every $n$, $$ \log(P_n)=\frac1n\sum\limits_{k=1}^nf\left(\frac{k}n\right), \qquad\text{with}\quad f(x)=\frac{\log(1+x)}x. $$ The function $f$ is continuous on $[0,1]$ (define $f(0)=1$) hence its Riemann sums converge to its integral and $P_n\to\mathrm e^\ell$ with $$ \ell=\int_0^1f(x)\mathrm dx=\int_0^1\left(\sum_{n\geqslant1}(-1)^{n+1}\frac{x^{n-1}}n\right)\mathrm dx=\sum_{n\geqslant1}\frac{(-1)^{n+1}}{n^2}=\frac{\pi^2}{12}. $$

Answer 2

The limit is $+\infty$: Let's call $P_n = \displaystyle\prod_{i=1}^n(1+\frac{i}{n})^{\frac{1}{i}}$, and $S_n=\ln P_n = \displaystyle\sum_{i=1}^n\frac{1}{i}\ln(1+\frac{i}{n})$.

Then you have $S_n-S_{[\frac{n}{2}]}=\displaystyle\sum_{i\geq\frac{n}{2}}^n\frac{1}{i}ln(1+\frac{i}{n})\geq[\frac{n}{2}]\times\frac{1}{n}\times\ln(1+\frac{\frac{n}{2}}{n})\rightarrow \frac{1}{2}\ln(\frac{3}{2})\neq 0$ which means that $S_n$ diverges (if $(S_n)$ had a limit , $S_n - S_{[\frac{n}{2}]}$ would go to $0$).