TITLE: Evaluating $\frac{1^2}{2^1}+\frac{3^2}{2^2}+\frac{5^2}{2^3}+\cdots$ using sigma notation
QUESTION [5 upvotes]: This question can be solved by method of difference
but I want to solve solve it using sigma notation:
$$\frac{1^2}{2^1}+\frac{3^2}{2^2}+\frac{5^2}{2^3}+\cdots+\frac{(2r +1)^2}{2^r}+\cdots$$
I used the geometric progression summation for the $(1/2)^r$ part, and opened $(2r+1)^2$ to $(4r^2 + 1 + 4r)$. If I now express this in sigma notation, it becomes
$$\frac{4n(n+1)(2n+1)}{6} + n + \frac{4(n\cdot n+1)}{2}$$ but I getting problem while putting upper limit infinity.
Where did I go wrong? Please explain.
Answer = $17$

REPLY [3 votes]: Consider$$S_r=\sum_{n=1}^r(2n-1)^{2}x^n=\sum_{n=1}^r({4n^{2}-4n+1})x^n$$ Now, rewrite
$$n^2=n(n-1)+n$$ which makes
$$S_r=\sum_{n=1}^r(4n(n-1)+1)x^n=4x^2\sum_{n=1}^r n(n-1)x^{n-2}+\sum_{n=1}^r x^n$$ that is to say $$S_r=4x^2 \left(\sum_{n=1}^r x^n \right)''+\sum_{n=1}^r x^n$$
Recall that $$\sum_{n=1}^r x^n=\frac{x \left(1-x^r\right)}{1-x}$$ Compute the second derivative and simplify; when finished, make $x=\frac 12$ and take the limit for $r \to \infty$.