Find the sum of $\sum _{n=1}^\infty n (2 n+1) x^{2 n}$ using only differentiation and knowing that $\sum _{n=0}^\infty x^n=\frac{1}{1-x}$
I started like that:
$n^2=n(n-1)+n$
so $\sum _{n=1}^\infty n(2n+1)x^{2n}=\sum _{n=1}^\infty (2n^2+n)x^{2n}=2\sum _{n=?}^\infty n(n-1)x^{2n}+3\sum _{n=?}^\infty nx^{2n}$
but as you can see I don't know at which $n$ does the summing start (that's why I marked it with "?").
Then I would go $x^2=t$
so $2\sum _{n=?}^\infty n(n-1)t^{n}+3\sum _{n=?}^\infty nt^{n} = 2t^2\sum _{n=?}^\infty n(n-1)t^{n-2}+3t\sum _{n=?}^\infty nt^{n-1} = 2t^2(\sum _{n=0?}^\infty t^n)''+3t(\sum _{n=0?}^\infty t^n)'=2t^2(\frac{1}{1-t})''+3t(\frac{1}{1-t})'$
and then it is simple.
But am I right? Can one calculate the sum like that? If no, how to find constant when integrating $\sum _{n=1}^\infty n(2n+1)x^{2n}$?
Because $\int \sum _{n=1}^\infty n(2n+1)x^{2n}dx=\sum _{n=1}^\infty n(2n+1)\frac{x^{2n+1}}{2n+1}+C$ and I don't know how to find the C if I don't know what function does represent our sum.