2

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.

user1176409
  • 1
B2K
  • 415
  • 1
    Expand the polynomial $(k+1)(k+3)$ as $(k+1)(k+2)+(k+1)$. Split the series in two. Also, take first derivatives of $\sum_{k=0}^{\infty}x^k$ and also second derivative. Look how those derivatives look when you replace $x=\frac{-1}{3}$. – logarithm May 15 '19 at 22:48

3 Answers3

6

$$\frac{1}{1+x}=\sum_{k=0}^{\infty}(-1)^kx^k$$ 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 $x=\frac{1}{3}$

E.H.E
  • 23,280
1

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 $0$

lab bhattacharjee
  • 274,582
1

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.

Claude Leibovici
  • 260,315