$$ \sum_0^\infty u^{2n+1} = \frac{u}{1-u^2} $$ $$ \Rightarrow \sum_0^\infty (2n+1)u^{2n} = \frac{d}{du}\left(\frac{u}{1-u^2}\right)= \frac{u^2+1}{(1-u^2)^2} $$ In our problem, we have $u= t^n(1-t^n)x^n$. But my professor has it written down that: $$ \sum_0^\infty (2n+1)u^{2n} = \frac{1-3u^2}{(1-u^2)^2} $$ We began with: $$ \sum_0^\infty (2n+1)u^{n}. $$ and changed it to: $$ \sum_0^\infty (2n+1)u^{\frac{2n}{2}} $$ I do not understand how my professor went from line 1 to line 3. The only thing I can see causing this is going from line 4 to 5 changing it. But I don't see how.
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It looks like your professor made a mistake. Your derivative is correct. His is wrong. I believe he flipped a sign when using the quotient rule (something like turning $\dots-u(-2u^2)$ into $\dots-u(2u^2)$. – Bill Cook Mar 20 '14 at 02:19
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Ah I see how he could've made this mistake. I will continue from here and see if my answer is correct. – H5159 Mar 20 '14 at 02:32
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@BillCook If we assume my differentiation is the correct method, and we substitute $\sqrt{u}^{2n}$ for $u$, does this mean in the closed form we replace $u$ with $\sqrt{u}$? – H5159 Mar 20 '14 at 02:37
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You can replace each $u$ with $\sqrt{u}$ and the identity will still hold (assuming $\sqrt{u}$ lies within the domain of convergence. – Bill Cook Mar 20 '14 at 02:39
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@BillCook When I use the derivative as we did, it does not output the correct answer. – H5159 Mar 20 '14 at 02:58
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I also think that your professor made a mistake. – Claude Leibovici Mar 20 '14 at 04:21
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He was speaking of another problem when I took notes and I got it mistaken. It is settled. – H5159 Mar 20 '14 at 15:22