$\lim\limits_{n\to\infty} \prod\limits_{k=1}^{n} \left( 1 + \tan{\frac{k}{n^2}} \right)$

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?

• In my humble opinion, this is the most natural solution we can think of. – Sangchul Lee Mar 15 '19 at 20:53
• In my humble opinion too, this is probably the simplest way to do it for the limit. Moreover, this allow to have very good approximations of the partial products. I shall put an answer for that. – Claude Leibovici Mar 16 '19 at 6:58

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.

• That's pretty nice; One thing: You don't need to use the integral for that limit. Notice that $\sum_{k=1}^{n} \frac{k}{n^2} = \frac{\sum_{k=1}^n k}{n^2} = \frac{n(n+1)}{2n^2}$ which clearly goes to $\frac{1}{2}$. – Davidmath7 Mar 15 '19 at 22:11
• @Davidmath7 oh, indeed, good catch! I had that integral in my mind for some unexplainable reasons ... – rtybase Mar 15 '19 at 22:15

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 $$\frac{3719 \sqrt{e}}{3600}\approx 1.70322$$