Series... Help...?

Question

Find the sum of the series
infinity sigma n=1 1/ [n(n+2)(n+4)]
I took the integral but I just could not get the answer.
The answer is 11/96.


What is the easiest way to prove that
(e^2x-1)/(e^2x+1)= x -1/3*x^3 + 2/15*x^5....

I had tried to take the partial fraction but it seems tat it just make the question more complicated because I have to do a lot of expansion. Is there any other method?

Answer 1

For e^(2x)-1 / e^(2x)+1:
that expression is equal to tanh x. The hyperbolic tangent.
Get the Taylor's expansion for tahn x centered at 0. ©
f(x) = tanh x
f'(x) = sech²(x)
f''(x) = -2sech²(x)tanh(x)
f'''(x) = 4sech²(x)tanh²(x) - 2sech⁴(x)
f^(4)(x) = ...

f(0)=0
f'(0)=1
f''(0)=0
f'''(0)=-2

f(x) = tanh x = 0 +1*x + 0 - 2/6*x³ + 0 - ...

Taylors expansion:
f(x) = f(a) + f'(a)(x-a)/1! + f''(a)(x-a)²/2! + f'''(a)(x-a)³/3! + f^(4)(a)(x-a)⁴/4!+...




Edit: I have professorbundles to thank for the next proof:
Σ[n=1..∞] (1/ [n(n+2)(n+4)]) .... by partial fraction
= Σ[n=1..∞] (1/[8n] - 1/[4(n+2)] + 1/[8(n+4)]) ... i'll not write the indices later. It will be understood.

now: 1/n = ∫₀¹ x^(n-1)dx , 1/(n+2) = ∫₀¹ x^(n+1)dx , 1/(n+4) = ∫₀¹ x^(n+3)dx

thus the summation becomes:
Σ ∫₀¹ [(1/8)x^(n-1) - (1/4)x^(n+1) + (1/8)x^(n+3)] dx
= ∫₀¹ { Σ x^n [1/(8x) - x/4 + x³/8] } dx .... note: Σ[n=1..∞] x^n = x/(1-x)
= ∫₀¹ x/(1-x) * [1/(8x) - x/4 + x³/8] dx
= 1/8 ∫₀¹ [1 - 2x² + x⁴] / (1-x) dx
= 1/8 ∫₀¹ [1 + x - x² - x³] dx
= 1/8 [x + x²/2 - x³/3 - x⁴/4]₀¹
= 1/8 [1 + 1/2 - 1/3 - 1/4]
= 11/96.