$\lim\limits_{n\to\infty} \prod\limits_{k=1}^{n} \left( 1 + \tan{\frac{k}{n^2}} \right) $ 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?
 A: 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.
A: Too long for a comment.
As I wrote in comments, composition of Taylor series is not only good for the limit but also allows quick and reasonable approximation of the partial product
$$P_n= \prod_{k=1}^{n} \left( 1 + \tan \left(\frac{k}{n^2}\right) \right)$$ Doing what you did taking logarithms and Taylor expansion, we have
$$\tan \left(\frac{k}{n^2}\right)=\frac{k}{n^2}+\frac{k^3}{3 n^6}+O\left(\frac{1}{n^{10}}\right)$$ making
$$\log \left(1+\tan \left(\frac{k}{n^2}\right)\right)=\frac{k}{n^2}-\frac{k^2}{2 n^4}+\frac{2 k^3}{3 n^6}+O\left(\frac{1}{n^{8}}\right)$$
$$\log(P_n)=\frac{1}{2}+\frac{1}{3 n}-\frac{1}{12 n^2}+\frac{2}{15 n^3}+O\left(\frac{1}{n^{4}}\right)$$ Continuing with Taylor, using $P_n=e^{\log(P_n)}$
$$P_n=\sqrt e \left(1+\frac{1}{3 n}-\frac{1}{36 n^2}+O\left(\frac{1}{n^{3}}\right) \right)$$
Computing for $n=10$, the exact result is $1.70341$ while the above approximation gives $\frac{3719 \sqrt{e}}{3600}\approx 1.70322$
