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Here's my question, it is rather straightforward: does $\sum_{n=0}^{\infty} \frac{(n+1)(n+2)}{2^n}=16$? First I tested if $\sum_{n=0}^{\infty} \frac{(n+1)(n+2)}{2^n}$ diverges to make sure it doesn't add to $\infty$. I found that taking the limit of the series as n approaches $\infty$ gives no information.

Then I expanded the series to get $\frac{2}{1}+\frac{6}{2}+\frac{12}{4}+....$ and found no pattern. My goal is to evaluate the series and find out if it's equal to 16 or not. How should I do that? or is there another approach to the question?

Metaphysics of Morals
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2 Answers2

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$$S:=\sum_{n=0}^{\infty} \frac{(n+1)(n+2)}{2^n}$$

$$\frac S2:=\sum_{n=0}^{\infty} \frac{(n+1)(n+2)}{2^{n+1}}=\sum_{n=0}^{\infty} \frac{n(n+1)}{2^n}$$

so that

$$S-\frac S2=\sum_{n=0}^{\infty} \frac{2(n+1)}{2^n}.$$

Then

$$T:=\sum_{n=0}^{\infty} \frac{n+1}{2^n}$$

and

$$T-\frac T2=\sum_{n=0}^{\infty} \frac1{2^n}.$$

  • Rhôo, this index shifting to $n=1$ readjusted to $n=0$ since $a_0=0$ is so pervert! I like the idea, though of not using derivation (even though it is finite differences in disguise) for once, this changes. – zwim Mar 31 '21 at 19:58
  • +1 I love this method, it's really ingenious! Thank you! – A-Level Student Mar 31 '21 at 21:49
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hint

For $ x $ such that $ |x|<1$, you might know that

$$F(x)=1+x+x^2+x^3+...=$$ $$\sum_{n=1}^\infty x^n=\frac{1}{1-x}$$

$$xF'(x)=x+2x^2+3x^3+...=$$ $$\sum_{n=1}^\infty nx^n=x\frac{1}{(1-x)^2}$$

differentiate again to get

$$x\frac{ d}{dx}(xF'(x))=$$ $$x+2^2x^2+3^2x^3+...=\sum_{n=1}^{\infty}n^2x^n$$

Now, observe that $$\frac{(n+1)(n+2)}{2^n}=$$ $$n^2\frac{1}{2^n}+3n\frac{1}{2^n}+2\frac{1}{2^n}=$$ $$n^2x^n+3nx^n+2x^n$$

hamam_Abdallah
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