A fine summation!

Calculus Level 4

2 0 1 + tan ⁡ 2 0 x + 2 1 1 + tan ⁡ 2 1 x + 2 2 1 + tan ⁡ 2 2 x + ⋯ + 2 n 1 + tan ⁡ 2 n x + ⋯ \dfrac{2^0}{1 + \tan^{2^0} x} + \dfrac{2^1}{1 + \tan^{2^1} x} + \dfrac{2^2} {1 + \tan^{2^2} x} + \cdots + \dfrac {2^n}{1 + \tan^{2^n} x } + \cdots

Evaluate the series above for x = π 3 x = \dfrac \pi3 .


The answer is 1.366.

Rajyawardhan Singh by Rajyawardhan Singh , 19, India

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2 solutions

Let me denote the above summation by S S

now adding 1 1 − t a n x \frac{1}{1-tanx} to both sides .

S S + 1 1 − t a n x \frac{1}{1-tanx} = 1 1 − t a n x \frac{1}{1-tanx} + 1 1 + t a n x \frac{1}{1+tanx} + 2 1 + t a n 2 x \frac{2}{1+tan^2x} + ...

S S + 1 1 − t a n x \frac{1}{1-tanx} = 2 1 − t a n 2 x \frac{2}{1-tan^2x} + 2 1 + t a n 2 x \frac{2}{1+tan^2x} + 4 1 + t a n 4 x \frac{4}{1+tan^4x} +...

S S + 1 1 − t a n x \frac{1}{1-tanx} = 4 1 − t a n 4 x \frac{4}{1-tan^4x} + 4 1 + t a n 4 x \frac{4}{1+tan^4x} + ...

And so on the r.h.s of the sum will get merging.

Thus S S = − 1 1 − t a n x \frac{-1}{1-tanx} = 1 t a n x − 1 \frac{1}{tanx-1}

Now putting x x = π 3 \frac{\pi}{3}

we get S S = 1 3 − 1 \frac{1}{\sqrt3 -1}

1 Helpful 1 Interesting 7 Brilliant 2 Confused
Kushal Dey
May 11, 2021

2/(1-x²)-1/(1-x)=1/(1+x)
4/(1-x⁴)-2/(1-x²)=2/(1+x²)
......
Adding all equations,
Lt n->infinity 2ⁿ/(1-x^(2ⁿ))-1/(1-x)=1/(1+x)....
Note that limit in lhs will be finite only if |x|>1.
We have to put x=tan(π/3) so we're good to go. Thus value of the series is 1/(sqrt(3)-1)
=(sqrt(3)+1)/2.



2 Helpful 2 Interesting 1 Brilliant 1 Confused

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