Sum to infinity series Consider the following series $$\sum_{k=0}^{\infty}\frac{(k+1)(k+3)(-1)^k}{3^k}$$
I'm told to analytically find the sum to infinity and I have been given this as a clue. 
$$\Sigma_{k=0}^\infty x^k = \frac{1}{1-x} \text{ if } |x|<1$$
I know that the answer is $\frac{45}{32}$ but that's just because of wolfram. I really have no idea where to begin. 
I tried to write it out and find a pattern but I could not spot one and I don't understand how the hint is useful here. 
 A: $$\frac{1}{1+x}=\sum_{k=0}^{\infty}(-1)^kx^k$$
multiply by $x$
$$\frac{x}{1+x}=\sum_{k=0}^{\infty}(-1)^kx^{k+1}$$
$$(\frac{x}{1+x})'=\sum_{k=0}^{\infty}(-1)^k(k+1)x^{k}$$
multiply by $x^3$
$$x^3(\frac{x}{1+x})'=\sum_{k=0}^{\infty}(-1)^k(k+1)x^{k+3}$$
$$(x^3(\frac{x}{1+x})')'=\sum_{k=0}^{\infty}(-1)^k(k+1)(k+3)x^{k+2}$$
divide by $x^2$
$$\frac{1}{x^2}(x^3(\frac{x}{1+x})')'=\sum_{k=0}^{\infty}(-1)^k(k+1)(k+3)x^{k}$$
$$\frac{(x+3)}{(x+1)^3}=\sum_{k=0}^{\infty}(-1)^k(k+1)(k+3)x^{k}$$
now let $x=\frac{1}{3}$
A: Let $(k+1)(k+3)(-1/3)^k=f(k+1)-f(k)$
so that $$\sum_{k=0}^m(k+1)(k+3)(-1/3)^k=f(m+1)-f(0)$$
where $f(n)=(-1/3)^n(a_0+a_1n+\cdots)$
$(k+1)(k+3)=a_0(1+3)+a_1(k+3(k-1))+a_2(?)+\cdots$
As the coefficients of $k^r$ is $0$ for $r\ge3$
$a_r=0$ for $r\ge3$
Compare the coefficients of $k^2,k,k^0$ to find $a_2,a_1,a_0$
Now set $m\to\infty$
We can use
https://en.m.wikipedia.org/wiki/Root_test
and Proof that $\lim_{n\rightarrow \infty} \sqrt[n]{n}=1$
to prove $\lim_{m\to\infty}f(m+1)$ converges to $0$
A: In the problem, you are given "nice" terms and factors.
Let us make it more general as
$$S=\sum_{k=0}^{\infty}\frac{(a k^2+b k+c)(-1)^k}{3^k}$$ and let $x=-\frac13$ to get
$$S=\sum_{k=0}^{\infty}(a k^2+b k+c)x^k$$ Now, the trick
$$k =(k-1)+1 \qquad \text{and} \qquad k^2=k(k-1)+(k-1)+1$$ makes
$$a k^2+b k+c=a [k(k-1)+(k-1)+1]+b[(k-1)+1]+c$$
$$a k^2+b k+c=a k(k-1)+(a+b)(k-1)+(a+b+c)$$
$$S=a \sum_{k=0}^{\infty} k(k-1) x^k+(a+b) \sum_{k=0}^{\infty}(k-1) x^k+(a+b+c)\sum_{k=0}^{\infty} x^k$$ that is to say
$$S=ax^2 \sum_{k=0}^{\infty} k(k-1) x^{k-2}+(a+b)x \sum_{k=0}^{\infty}(k-1) x^{k-1}+(a+b+c)\sum_{k=0}^{\infty} x^k$$ that is to say
$$S=a x^2 \left(\sum_{k=0}^{\infty} x^k \right)''+(a+b)x\left(\sum_{k=0}^{\infty} x^k \right)'+(a+b+c)\left(\sum_{k=0}^{\infty} x^k \right)$$ Now, using the hint, it is quite simple.
