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Finding sum of series $\displaystyle \sum^{\infty}_{k=0}\frac{(k+1)(k+2)}{2^k}$

what i try

Let $\displaystyle x=\frac{1}{2},$ Then series sum is $\displaystyle \sum^{\infty}_{k=0}(k+1)(k+2)x^k$

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

differentiating both side

$\displaystyle \sum^{\infty}_{k=0}(k+1)x^{k}=+\frac{1}{(1-x)^2}$

How do i solve it help me please

David C. Ullrich
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jacky
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2 Answers2

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You almost have it. Instead of multiplying the infinite sum by $x$, try multiplying by $x^2$, so you get

$$\sum^{\infty}_{k=0}x^{k+2} = \frac{x^2}{1-x} \tag{1}\label{eq1A}$$

After differentiating twice, you will then get your series sum on the left, i.e., $\sum^{\infty}_{k=0}(k + 1)(k + 2)x^{k}$. You can determine the right side at $x = \frac{1}{2}$. I'll leave the details to you to finish.

John Omielan
  • 47,976
4

Another way:

Let $f(m)=\dfrac{a+bm+cm^2}{2^m}$

and $\dfrac{(k+1)(k+2)}{2^k}=f(k)-f(k+1)$

so that $$\sum_{k=0}^n(f(k)-f(k+1))=f(0)-f(n+1)$$

$$\implies2(k+1)(k+2)=2(a+bk+ck^2)-[a+b(k+1)+c(k+1)^2]$$

$$\iff4+6k+2k^2=k^2(2c-c)+k(2b-b-2c)+2a-a-b-c$$

Compare the constants and the coefficients of $k,k^2$ to find $a,b,c$

Now we can prove $$\lim_{n\to\infty}f(n+1)=0$$

lab bhattacharjee
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