# Thread: Sum of the sereis

1. ## Sum of the sereis

Helllo, frnds,

S(n) = (1/2) + (2^2)/2^2 + (3^2)/2^3 + (4^2)/2^4 + .........

So, we can write this series as

Sum of T(n) = [(n^2)/(2^n)] ...

Thanx..

2. The placement of this question in an elementary mathematics forum is puzzling.
This requires a bit of calculus at a higher level to solve.
Questions: Do you simply want a answer for some simple reason?
Or are you required to supply reasons for the particular answer?

Please respond so that we know how to answer and to what extent the answer should be.

3. Hello,

I really apologize for the wrong placement of this question. I want the full answer with reasons. Shall I open a new thread in the relevant section?

Thanks

4. Hint 1: Evaluate $\sum_{n=1}^{\infty} n^2 x^n$ by differenciating term-by-term the expression $\sum_{n=0}^{\infty}x^n = \frac{1}{1-x}$ twice.

Hint 2: Now consider $\sum_{n=1}^{\infty} \frac{n^2x^n}{2^n} = \sum_{n=1}^{\infty} n^2 \left( \frac{x}{2} \right)^n$ and evaluate this sum using #1.

Hint 3: Let $x=1$ to get your answer.

5. $\begin{array}{l}
f(r) = \sum\limits_{n = 1}^\infty {r^n } = \frac{r}{{1 - r}},\quad \left| r \right| < 1 \\
g(r) = rf'(r) = \sum\limits_{n = 1}^\infty {nr^n } = \frac{r}{{\left( {1 - r} \right)^2 }} \\
rg'(r) = \sum\limits_{n = 1}^\infty {n^2 r^n } = \color{red}{\frac{{r\left( {1 + r} \right)}}{{\left( {1 - r} \right)^3 }}} \\
\end{array}$

Now let $r=\frac{1}{2}$

6. Thanx a lot for your help!!!