Sum of power series $\sum_{n=1}^\infty (-1)^n\frac{n(n+1)}{2^n}x^n$

Question

Calculate the sum of series:

$$\sum_{n=1}^\infty (-1)^n\frac{n(n+1)}{2^n}x^n$$

I know how to calculate sum of power series, but I don't know what should I do with $(-1)^n$

Answer 1

Hint:

Try first to find what is $$ \sum_{n=1}^\infty n (n+1)y^n $$ by noting that $$ n(n+1)y^n=(y^{n+1})''\cdot y $$

A most important technique about calculation of power series is differentiation (and integration) term by term, which should be discussed in any serious real (complex) analysis textbook. See also a note by Gowers.

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Besides the "formal" calculation, another issue you still need to address is that for what $y$ the series is convergent.

Answer 2

HINT:

$$\sum_{n=1}^\infty (-1)^n\frac{n(n+1)}{2^n}x^n=\sum_{n=1}^\infty n^2\left(-\dfrac12\right)^n+\sum_{n=1}^\infty n\left(-\dfrac12\right)^n$$

Now for $|y|<1,$ $$\sum_{r=0}^\infty y^r=\dfrac1{1-y}$$

Differentiate both sides wrt $y$ $$\sum_{r=1}^\infty ry^{r-1}=\dfrac1{(1-y)^2}$$

$$\sum_{r=1}^\infty ry^r=\dfrac y{(1-y)^2}=\dfrac{1-(1-y)}{(1-y)^2}=\dfrac1{(1-y)^2}-\dfrac1{1-y}$$

Differentiate both sides wrt $y$ and multiply by $y$

Answer 3

$$\frac{1}{1+x}=\sum _{n=0}^{\infty }(-1)^{n}x^n$$ $$\frac{-x}{1+x}=\sum _{n=1}^{\infty }(-1)^{n}x^n$$

multiply by $x$ $$\frac{-x^2}{1+x}=\sum _{n=1}^{\infty }(-1)^{n}x^{n+1}$$ $$(\frac{-x^2}{1+x})''=\sum _{n=1}^{\infty }(-1)^{n}n(n+1)x^{n-1}$$ then multiply by $x$ and let $x\rightarrow \frac{x}{2}$

Answer 4

Let $y=-\dfrac{x}{2}$, so we have to find the power series $$\sum_{r=1}^{\infty}{r(r+1)y^r}$$

Note that $$\sum_{r=1}^{\infty}{y^r}=\dfrac{1}{1-y}$$ On differentiating throughout twice we arrive at the following

$$\sum_{r=1}^{\infty}{r(r-1)y^{r-2}}=\dfrac{2y}{(1-y)^3}$$

From here onward it is pretty straightforward if you assign $r=n+1$ and substitute $y=-\dfrac{x}{2}$.

It is also to be noted that for the power series to converge $|y|\lt 1\implies \left|{x}\right|\lt 2$