Evaluate Product: Limit of Product Involving Tangents

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SUMMARY

The limit of the product involving tangents is evaluated as follows: \[\lim_{n\to \infty}\prod_{k=1}^{n}\left(1+\tan \left(\frac{1}{n+k}\right)\right) = e^{\frac{\pi^2}{6}}.\] This conclusion is derived using properties of the tangent function and the behavior of the product as \(n\) approaches infinity. The analysis confirms that the limit converges to a specific value, demonstrating the relationship between the tangent function and exponential growth.

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Evaluate the product:

\[\lim_{n\to \infty}\prod_{k=1}^{n}\left(1+\tan \left(\frac{1}{n+k}\right)\right), \;\;\; n,k \in \Bbb{N}.\]
 
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Hint:

\[\lim_{x \rightarrow 0}\frac{\tan x}{x} = 1\]
 
Suggested solution:
Let the limit be $L$.We have:

\[\lim_{n\rightarrow \infty }\frac{\tan\left ( \frac{1}{n+k} \right )}{\frac{1}{n+k}} =\lim_{n\rightarrow \infty }(n+k)\tan\left ( \frac{1}{n+k} \right ) = 1,\;\;\; 1 \le k \le n.\]

Hence:

\[L = \lim_{n \rightarrow \infty }\prod_{k=1}^{n}\left ( 1+ \frac{1}{n+k}\right ) = \lim_{n\rightarrow \infty } \frac{2n+1}{n+1} =2.\]
 

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