I need to find the limit of the following sequence: $$\lim\limits_{n \to\infty}\prod_{k=1}^n \left(1+\frac{k}{n^2}\right)$$


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In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities

$$\bbox[5px,border:2px solid #C0A000]{\frac{x-1}{x}\le \log(x)\le x-1} \tag 1$$

for $x>0$.

Note that we have

$$\begin{align} \log\left(\prod_{k=1}^n \left(1+\frac{k}{n^2}\right)\right)&=\sum_{k=1}^n \log\left(1+\frac{k}{n^2}\right)\tag 2 \end{align}$$

Applying the right-hand side inequality in $(1)$ to $(2)$ reveals

$$\begin{align} \sum_{k=1}^n \log\left(1+\frac{k}{n^2}\right)&\le \sum_{k=1}^n \frac{k}{n^2}\\\\ &=\frac{n(n+1)}{2n^2} \\\\ &=\frac12 +\frac{1}{2n}\tag 3 \end{align}$$

Applying the left-hand side inequality in $(1)$ to $(2)$ reveals

$$\begin{align} \sum_{k=1}^n \log\left(1+\frac{k}{n^2}\right)&\ge \sum_{k=1}^n \frac{k}{k+n^2}\\\\ &\ge \sum_{k=1}^n \frac{k}{n+n^2}\\\\ &=\frac{n(n+1)}{2(n^2+n)} \\\\ &=\frac12 \tag 4 \end{align}$$

Putting $(2)-(4)$ together yields

$$\frac12 \le \log\left(\prod_{k=1}^n \left(1+\frac{k}{n^2}\right)\right)\le \frac12+\frac{1}{2n} \tag 5$$

whereby application of the squeeze theorem to $(5)$ gives

$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty} \log\left(\prod_{k=1}^n \left(1+\frac{k}{n^2}\right)\right)=\frac12}$$

Hence, we find that

$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to \infty}\prod_{k=1}^n \left(1+\frac{k}{n^2}\right)=\sqrt e}$$

And we are done!

  • $\begingroup$ That solution would be perfect for me if I knew anything about that 'O' symbol. My lecturer has not defined it yet, so there comes my question: is there any slightly less difficult solution? However, thank you for that one. $\endgroup$ – R.K. Dec 10 '16 at 16:33
  • $\begingroup$ Have you learned Taylor's Theorem? $\endgroup$ – Mark Viola Dec 10 '16 at 16:36
  • $\begingroup$ Unfortunately, not yet. I'm just at the beginning of my studies. $\endgroup$ – R.K. Dec 10 '16 at 16:38
  • $\begingroup$ @R.K. I've edited the development given your feedback. The development relies on basic tools only now. $\endgroup$ – Mark Viola Dec 10 '16 at 17:30
  • 1
    $\begingroup$ @marcocantarini Marco, thanks. And once in a while, I'll bring heavIer weaponry $\endgroup$ – Mark Viola Dec 10 '16 at 22:37

Note that $$\prod_{k\leq n}\left(1+\frac{k}{n^{2}}\right)=\prod_{k\leq n}\left(\frac{n^{2}+k}{n^{2}}\right)=\frac{\left(n^{2}+1\right)_{n}}{n^{2n}} $$ where $\left(x\right)_{m} $ is the Pochhammer symbol and since $$\left(x\right)_{m}=\frac{\Gamma\left(x+m\right)}{\Gamma\left(x\right)} $$ we have $$\prod_{k\leq n}\left(1+\frac{k}{n^{2}}\right)=\frac{\Gamma\left(n^{2}+n+1\right)}{n^{2n}\Gamma\left(n^{2}+1\right)}\longrightarrow\color{red}{\sqrt{e}} $$ where the last limit can be calculated using the Stirling's approximation.

  • 4
    $\begingroup$ Bringing a gun to a knife fight. (+1) $\endgroup$ – Mark Viola Dec 10 '16 at 18:34
  • $\begingroup$ @Dr.MV Well, in a battle of the fittest, the gun usually wins... unless you are some amazing hero that saves us all :D $\endgroup$ – Simply Beautiful Art Dec 10 '16 at 19:51


$$P_n=\prod_{k=1}^n (1+\frac{k}{n^2})\implies \log(P_n)=\sum_{k=1}^n \log\left(1+\frac{k}{n^2}\right)$$ Work with the sum (without forgetting by the end that $P_n=e^{\log(P_n)}$)

  • $\begingroup$ Allo Claude! Comment allez vous? (+1) $\endgroup$ – Mark Viola Dec 10 '16 at 16:22

$$\left(1+\frac{k}{n^2} \right)\left(1+\frac{n-k}{n^2} \right)=\left(1+\frac{1}{n} +\frac{k(n-k)}{n^4} \right)$$

Then we have $$\left(1+\frac{1}{n} \right) \leq \left(1+\frac{1}{n} +\frac{k(n-k)}{n^4} \right) \leq \left(1+\frac{1}{n} +\frac{1}{n^2} \right)=\left(1 +\frac{n+1}{n^2} \right)$$

Therefore $$\left(1+\frac{1}{n} \right) ^\frac{n}{2} \leq \prod_{k=1}^n\left(1+\frac{k}{n^2}\right) \leq \left(1 +\frac{n+1}{n^2} \right)^\frac{n}{2}$$

and squeeze it.

  • $\begingroup$ @Dr.MV Forgot the product sign, fixed, ty. $\endgroup$ – N. S. Dec 10 '16 at 18:05

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