I want to calculate lim
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?
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.
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