How to evaluate this limit with Sandwich Theorem

Question

Here it is the limit \begin{align*} \lim_{n\rightarrow\infty}\prod_{k=1}^{n}\left(1 + \frac{k}{n^{2}}\right) \end{align*}

I have proved that it is smaller than the square root of $e$

Answer 1

$$\prod_{k=1}^n (1 + \frac{k}{n^2}) = \exp(\ln(\prod_{k=1}^n (1+\frac{k}{n^2})) = \exp(\sum_{k=1}^n\ln(1+\frac{k}{n^2})) $$

We'll deal with inner sum. Note that forall small $\epsilon$ we can take $M(\epsilon)$ big enough, such that for $n \ge M(\epsilon)$, we have $\frac{1}{n} < \epsilon$, so $\frac{k}{n^2} \le \frac{1}{n} < \epsilon$, for any $k \in \{1,...,n\}$, so it is a good idea to expand function $x \to \ln(1+x)$ near $0$. We have $\ln(1+\frac{k}{n^2}) = \frac{k}{n^2} - \frac{k^2}{2n^4} + o(\frac{1}{n^2})$ (that $\frac{1}{n^2}$, due to $\frac{k^2}{n^4} \le \frac{1}{n^2}$).

So our sum is: (I'll split it to 3 sums, if all of them will have limit, then our sum, too)

$$ \frac{1}{n}\sum_{k=1}^n \frac{k}{n} - \frac{1}{2n^4}\sum_{k=1}^{n}k^2 + n\cdot o(\frac{1}{n^2})$$

The last term goes to $0$, the second one, too, because $\sum_{k=1}^n k^2 \le n^3$, so we have to deal with the first sum.

But it is a riemann sum, hence it converges too $\int_0^1 xdx = \frac{1}{2}$

So, we proved: $$\sum_{k=1}^n \ln(1+\frac{k}{n^2}) = \frac{1}{2}$$

So, $$ \prod_{k=1}^n (1+\frac{k}{n^2}) = \exp(\frac{1}{2}) $$

Answer 2

Another (similar) way to do it.

$$a_n=\prod_{k=1}^{n}\left(1 + \frac{k}{n^{2}}\right)\implies \log(a_n)=\sum_{k=1}^{n}\log\left(1 + \frac{k}{n^{2}}\right)$$

Now, using Taylor expansion $$\log\left(1 + \frac{k}{n^{2}}\right)=\frac{k}{n^2}-\frac{k^2}{2 n^4}+O\left(\frac{1}{n^6}\right)$$ Neglecting the higher order terms and using Faulhaber's formulae

$$\log(a_n)=\frac{1}{n^2}\sum_{k=1}^{n} k-\frac{1}{2 n^4}\sum_{k=1}^{n} k^2=\frac{(n+1) \left(6 n^2-2 n-1\right)}{12 n^3}$$which shows the limit of $\frac 12$ when $n\to \infty$ and then the limit for $a_n$.

We could even continue; using long division $$\log(a_n)=\frac{1}{2}+\frac{1}{3 n}-\frac{1}{4 n^2}+O\left(\frac{1}{n^3}\right)$$ and continuing with Taylor series $$a_n=e^{\log(a_n)}=\sqrt e \left(1+\frac{1}{3 n}-\frac{7}{36 n^2}+O\left(\frac{1}{n^3}\right)\right)$$ which shows the limit and how it is approached.

Moreover, this gives a good approximation. For example $a_{10}=\frac{26590959965797227}{15625000000000000}\approx 1.70182$ while the above truncated expression gives $\frac{3713 \sqrt{e}}{3600}\approx 1.70047$.