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?

  • $\begingroup$ In my humble opinion, this is the most natural solution we can think of. $\endgroup$ – Sangchul Lee Mar 15 at 20:53
  • $\begingroup$ 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. $\endgroup$ – Claude Leibovici Mar 16 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.

  • 1
    $\begingroup$ 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 $\frac{1}{2}$. $\endgroup$ – Davidmath7 Mar 15 at 22:11
  • $\begingroup$ @Davidmath7 oh, indeed, good catch! I had that integral in my mind for some unexplainable reasons ... $\endgroup$ – rtybase Mar 15 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 $\frac{3719 \sqrt{e}}{3600}\approx 1.70322$


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