Question

How to calculate sum of this? `sum_(n=1)^oo(-1)^n n(n-1)x^n`

Answer 1

See below.

Explanation:

Considering `abs x < 1`

`sum_(n=1)^oo(-1)^n n(n-1)x^n = x^2 d^2/(dx^2) sum_(n=1)^oo (-x)^n`

but `sum_(n=1)^oo (-x)^n = 1/(1-(-x))-1` and

`d^2/(dx^2) sum_(n=1)^oo (-x)^n = 2/(x+1)^3` then

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

Answer 2

`sum_(n=1)^oo(-1)^n n(n-1)x^n=(2x^2)/(1+x)^3` when `|x|<1`

Explanation:

We begin by writing out some of the coefficients:
`sum_(n=1)^oo(-1)^n n(n-1)x^n=2x^2-6x^3+12x^4-20x^5...=`

The first thing we want to look at is the coefficients (the degree of `x` can be quite easily adjusted by multiplying and dividing the series by `x`, so they aren't as important). We see that they all are multiples of two, so we can bring out a factor of two:
`=2(x^2-3x^3+6x^4-10x^5...)`

The coefficients inside this parenthesis can be recognized as the binomial series with a power of `alpha=-3`:
`(1+x)^alpha=1+alphax+(alpha(alpha-1))/(2!)x^2+(alpha(alpha-1)(alpha-2))/(3!)x^3...`

`(1+x)^-3=1-3x+6x^2-10x^3...`

We notice that the exponents of all the terms in the parenthesis are bigger by two compared to the series we just derived, so we must multiply `x^2` to get the right series:
`2x^2(1+x)^-3=2x^2-6x^3+12x^4-20x^5...`

This means that our series is (when it converges) equal to:
`(2x^2)/(1+x)^3`

Just to verify that we didn't make a mistake, we can quickly use the Binomial Series to compute a series for `2x^2(1+x)^-3`:
`2x^2(1+x)^-3=2x^2(1-3x+((-3)(-4))/(2!)x^2+((-3)(-4)(-5))/(3!)x^3...)=`

`=2x^2(1-3x+(4!)/(2*2!)x^2-(5!)/(2*3!)x^3...)=`

`=2x^2(1-3x+(4*3)/2x^2-(5*4)/2x^3...)=`

We can describe this pattern like so:
`=2x^2sum_(n=0)^oo (-1)^n(n(n-1))/2x^(n-2)=sum_(n=0)^oo(-1)^n n(n-1)x^n`

Since the first term is just `0`, we can write:
`sum_(n=1)^oo (-1)^n n(n-1)x^n`
which is the series we started with, verifying our result.

Now we just need to find out the convergence interval, to see when the series actually has a value. We can do this by looking at the convergence conditions for the binomial series and find that the series converges when `|x|<1`