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

Question

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

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Answer 1 · 2018-06-22T11:08:33

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

#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#