Question

Evaluate the taylor series sum: `sum_{n=2}^{∞} {(-1)^n 3^{n-1}(2n)}/(2^{2n-1})` ?

`sum_{n=2}^{∞} {(-1)^n 3^{n-1}(2n)}/(2^{2n-1})` use taylor series to evaluate this sum.

Answer 1

`sum_(n=2)^oo (-1)^n ((2n)3^(n-1))/2^(2n-1) = 33/49`

Explanation:

Start from the geometric series:

`sum_(n=0)^oo q^n = 1/(1-q)` for `abs q < 1`

Let `q=-x` to have:

`sum_(n=0)^oo (-1)^nx^n = 1/(1+x)` for `abs x < 1`

Differentiate term by term:

`sum_(n=1)^oo (-1)^n nx^(n-1) = -1/(1+x)^2` for `abs x < 1`

Let `x=3/4`:

`sum_(n=1)^oo (-1)^n n(3/4)^(n-1) = -1/(1+3/4)^2`

`sum_(n=1)^oo (-1)^n (n3^(n-1))/2^(2n-2) = -16/49`

`sum_(n=1)^oo (-1)^n ((2n)3^(n-1))/2^(2n-1) = -16/49`

Extract the term for `n=1`:

`-1+sum_(n=2)^oo (-1)^n ((2n)3^(n-1))/2^(2n-1) = -16/49`

`sum_(n=2)^oo (-1)^n ((2n)3^(n-1))/2^(2n-1) = -16/49+1`

`sum_(n=2)^oo (-1)^n ((2n)3^(n-1))/2^(2n-1) = 33/49`