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We briefly covered generating functions in class and most of the situations we covered we were given a recurrence to find a generating function for. I haven't gotten very far but I do believe $$G_a(x) = \sum_{n=2}^\infty({x^n\over(n-1)(n+1)})$$ and i think i will have to differentiate at some point but I am lost

L. Johnson
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Hint: If we rewrite the $n$-th term as $$\frac{1}{2}\cdot \frac{x^n}{n-1}-\frac{1}{2}\cdot \frac{x^n}{n+1},$$ things may look more familiar.

André Nicolas
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  • I saw an example that has just the right side of the part above, and it pulled a $1\over{x}$ out to get $1/{x} * {x^{n+1}\over{n+1}}$ and then it differentiated to get get $x_n$ and which is $1\over{1-x}$ and then it integrated to get ${-ln(1-x)+c}\over{x}$ I hope i explained the example well enough, but i am trying to apply it to this problem and run into some problems – L. Johnson Mar 09 '16 at 17:04
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    Yes, you are close, since $-\ln(1-x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\cdots$. For sum of the first terms in my answer, you need to multiply by $\frac{x}{2}$. A similar adjustment works for the second terms, except the sum starts a little late, so some subtraction is needed. – André Nicolas Mar 09 '16 at 17:11