I want to calculate $$\lim\limits_{n\to\infty} \prod_{k=1}^{n} \left( 1 + \tan{\frac{k}{n^2}} \right) $$
Taking logarithms, it's enough to find
$$\lim\limits_{n\to\infty} \sum_{k=1}^{n} \log\left( 1 + \tan{\frac{k}{n^2}} \right).$$
Since $\lim\limits_{n\to\infty} \tan{\frac{x}{n^2}} = 0$, we can combine the Taylor series near $0$ of $\log(1+x)$ with the taylor series of $\tan{x}$ near $0$ to obtain the limit $e^\frac{1}{2}$.
My question is: is there any nicer way of evaluating this limit?
Best Answer
Probably not nicer, but still a different way is to use the facts that $$\lim\limits_{x\rightarrow0}\frac{\tan{x}}{x}=1$$ and, as shown here $$\lim\limits_{n\rightarrow\infty} \frac{1}{n}\sum\limits_{k=1}^n f\left(\frac{k}{n}\right)= \int\limits_{0}^{1} f(x)dx$$
$$\sum_{k=1}^{n} \log\left( 1 + \tan{\frac{k}{n^2}} \right)= \sum_{k=1}^{n} \tan{\frac{k}{n^2}} \cdot \log\left( 1 + \tan{\frac{k}{n^2}} \right)^{\frac{1}{\tan{\frac{k}{n^2}} }}=\\ \sum_{k=1}^{n} \frac{k}{n^2}\cdot \color{red}{ \frac{\tan{\frac{k}{n^2}}}{\frac{k}{n^2}} \cdot \log\left( 1 + \tan{\frac{k}{n^2}} \right)^{\frac{1}{\tan{\frac{k}{n^2}} }}}$$ Because the part in red $\rightarrow 1$ when $n\rightarrow\infty$ for any $k=1..n$, using the definition of the limit, $\forall \varepsilon, \exists N(\varepsilon)$ s.t. $\forall n > N(\varepsilon)$ $$(1-\varepsilon)\left(\sum_{k=1}^{n} \frac{k}{n^2}\right)<\sum_{k=1}^{n} \log\left( 1 + \tan{\frac{k}{n^2}} \right)<(1+\varepsilon)\left(\sum_{k=1}^{n} \frac{k}{n^2}\right)$$ leading to $$\lim\limits_{n\rightarrow\infty}\sum_{k=1}^{n} \log\left( 1 + \tan{\frac{k}{n^2}} \right)= \lim\limits_{n\rightarrow\infty}\sum_{k=1}^{n} \frac{k}{n^2}$$ But then $$\lim\limits_{n\rightarrow\infty}\sum_{k=1}^{n} \frac{k}{n^2}= \lim\limits_{n\rightarrow\infty}\frac{1}{n}\left(\sum_{k=1}^{n} \frac{k}{n}\right)=\int\limits_{0}^{1}x dx =\frac{1}{2}$$ and the result follows.