MHB Evaluate Product: Limit of Product Involving Tangents

  • Thread starter Thread starter lfdahl
  • Start date Start date
  • Tags Tags
    Product
lfdahl
Gold Member
MHB
Messages
747
Reaction score
0
Evaluate the product:

\[\lim_{n\to \infty}\prod_{k=1}^{n}\left(1+\tan \left(\frac{1}{n+k}\right)\right), \;\;\; n,k \in \Bbb{N}.\]
 
Last edited:
Mathematics news on Phys.org
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:

\[L = \lim_{n \rightarrow \infty }\prod_{k=1}^{n}\left ( 1+ \frac{1}{n+k}\right ) = \lim_{n\rightarrow \infty } \frac{2n+1}{n+1} =2.\]
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

Similar threads

Replies
2
Views
2K
Replies
2
Views
1K
Replies
3
Views
2K
Replies
2
Views
1K
Replies
4
Views
1K
Back
Top