Tuesday, June 18, 2019

Limit of a sequence including infinite product. $limlimits_{n toinfty}prod_{k=1}^n left(1+frac{k}{n^2}right)$





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


Answer




PRIMER:



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!


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