$\lim_{n \to \infty} \sqrt[k]{\prod_{k=1}^n \left(1+ \frac{k}{n}\right)}$

Question

I was studying for my Calculus test and I've got stuck in the following activity:

Compute $$\lim_{n \to \infty} \sqrt[k]{\prod_{k=1}^n \left(1+ \frac{k}{n}\right)}$$

I've tried to solve it using limits of Riemann sums of the logarithm of the expression:

$$\begin{align}\lim_{n \to \infty} \log{{\prod_{k=1}^n \left(1+ \frac{k}{n}\right)^{1/k}}}&=\lim_{n \to \infty} \sum_{k=1}^n \frac{1}{k} \log{\left(1+\frac{k}{n}\right)}\\ &=\lim_{n \to \infty} \sum_{k=1}^n \frac{1}{n}\frac{1}{k/n} \log{\left(1+\frac{k}{n}\right)}\end{align}$$

Defining $f(x) = \frac{1}{x} \log{(1+x)}$, the limit can be writen like a Riemann Sum:

$$\begin{align}\lim_{n \to \infty} \log{{\prod_{k=1}^n \left(1+ \frac{k}{n}\right)^{1/k}}}&= \lim_{n \to \infty} \frac{1}{n}f\left(\frac{k}{n}\right)\\&= \int_{0}^1 \frac{1}{x} \log{(1+x)} \end{align}$$

However, I'm afraid this integral cannot be computed in terms of elementary functions. Can anyone see my error or give me another idea of doing the problem?

Edit

As Martin R have said, it's probably a typo so I'm going to propose the solution to the "more reasonable" problem and I would appreciate if anyone can tell me if it's correct. I would proceed analogously, calculating the limit of the logarithm:

$$\lim_{n \to \infty} \log{{\prod_{k=1}^n \left(1+ \frac{k}{n}\right)^{1/n}}}=\lim_{n \to \infty} \sum_{k=1}^n \frac{1}{n} \log{\left(1+\frac{k}{n}\right)}$$

Then, defining $f(x) = \log(1+x)$, we get that:

$$\begin{align}\lim_{n \to \infty} \log{{\prod_{k=1}^n \left(1+ \frac{k}{n}\right)^{1/n}}}&= \lim_{n \to \infty} \frac{1}{n} f\left(\frac{k}{n}\right) \\&= \int_0^1 \log(1+x)\end{align} $$

Integrating by parts, we get that one of its primitives is $\int \log(1+x) = (x+1) \log (1+x) - x$. Applying Barrow's Rule, we conclude that $\int_0^1 \log(1+x) = 2 \log 2 - 1$. Hence:

$$\begin{align}\lim_{n \to \infty} \sqrt[n]{\prod_{k=1}^n \left(1+ \frac{k}{n}\right)}& = e^{\lim_{n \to \infty} \log{{\prod_{k=1}^n \left(1+ \frac{k}{n}\right)^{1/n}}}}\\&= e^{2 \log 2 - 1} \\&= \frac{4}{e}\end{align}$$

Answer 1

Your technique works, and is quite direct.

Alternatively, $$1+\frac{k}{n}=\frac{n+k}{n}.$$

So your expression can be written as $$\sqrt[n]{\frac{(2n)!}{n^nn!}}$$ Then use Stirling’s approximation for $(2n)!$ and $n!$

Specifically, $$\frac{(2n)!}{n!n^n}\sim\frac{\sqrt{4\pi n}(2n/e)^{2n}}{\sqrt{2\pi n}(n/e)^{n}n^n}=\sqrt{2}\frac{2^{2n}}{e^n}$$

So the $n$th root goes to $\frac4e.$

Answer 2

I think it was supposed to be the product $$\displaystyle \mathcal{P} = \lim_{n \to \infty} {\prod_{k=1}^n \sqrt[k]{\left(1+ \frac{k}{n}\right)}}$$

We indeed arrive at the integral given by the OP:

\begin{align} \mathcal {\log P} &=\int_0^1\frac1x\log(1+x)\,\mathrm{d}x\\ &=\int_0^1\sum_{k=0}^\infty(-1)^k\frac{x^{k+1}}{k+1}\,\mathrm{d}x\\ &=\sum_{k=0}^\infty\frac{(-1)^k}{(k+1)(k+2)} \\& = \sum_{k=0}^\infty\frac{(-1)^k}{k+1}-\sum_{k=0}^\infty\frac{(-1)^k}{k+2}\\& = \log 2 - (1-\log 2) \\& = \log 4-1 \end{align}

So $\mathcal{P} = \exp(\log 4 -1) =4e^{-1}.$