Why $\sum_{i=0}^\infty i^2p^{i-1} = \sum_{i=0}^\infty i(\frac{d}{dp}p^{i-1})$?

Question

I'm going through a proof right now and am having trouble figuring out the math behind one line. It says:

$$\sum_{i=0}^\infty i^2p^{i-1} = \sum_{i=0}^\infty i(\frac{d}{dp}p^i) $$

I know this question is vague but can anybody explain why this is the case?

Use the calculus rule $\frac{d}{dx} x^i = i \frac{d}{dx} x^{i-1}$ with $x=p$. — angryavian, Jun 09 '17 at 22:46

score 4 · Accepted Answer · answered Jun 09 '17 at 22:47

4

First of all, the summation notation is unnecessary and irrelevant. Remove it and we get $$i^2p^{i-1}=i\big(\frac{d}{dp}p^i\big)$$ Okay. Recall the formula $$\frac{d}{dx}x^k=kx^{k-1}, k\ne0$$ Then we have that $$\frac{d}{dp}p^{i}=ip^{i-1}, i\ne0$$ and so, by substitution, we have $$i\big(\frac{d}{dp}p^i\big)=i\big(\frac{d}{dp}p^i\big)$$ $$i\big(\frac{d}{dp}p^i\big)=i(ip^{i-1})$$ $$i\big(\frac{d}{dp}p^i\big)=i^2p^{i-1}$$ Does that answer your question?

answered Jun 09 '17 at 22:47

Franklin Pezzuti Dyer

39,754
9
73
166

Thanks, that makes sense! Math rocks! – MarksCode Jun 09 '17 at 23:02
Haha! Indeed it does. – Franklin Pezzuti Dyer Jun 09 '17 at 23:03

Why $\sum_{i=0}^\infty i^2p^{i-1} = \sum_{i=0}^\infty i(\frac{d}{dp}p^{i-1})$?

1 Answers1