I'm reading this book about algorithms and data structures. In the exercise section of chapter 1, there is this sum :
$\displaystyle\sum_{i=0} ^{ \infty} \frac{i^{2}}{4^{i}}$
which we need to evaluate. my solution is the following:
//$S_n$ and $s_t$ are just names
$let: s_n =\displaystyle\sum_{i=0} ^{\infty} \frac{i^{2}}{4^{i}}$
$s_n = 0+\frac{1^{2}}{4^{1}}+\frac{2^{2}}{4^{2}}+\frac{3^{2}}{4^{3}}+\frac{4^{2}}{4^{4}}+\frac{5^{2}}{4^{5}}+....$
$\frac{1}{4}\times s_n = 0+\frac{1^{2}}{4^{2}}+\frac{2^{2}}{4^{3}}+\frac{3^{2}}{4^{4}}+\frac{4^{2}}{4^{5}}+....$
$ s_n - \frac{1}{4}s_n =\frac{1}{4}+\frac{3}{4^{2}}+\frac{5}{4^{3}}+\frac{7}{4^{4}}+....$
$let: s_t = \frac{3}{4^{2}}+\frac{5}{4^{3}}+\frac{7}{4^{4}}+....$
$there for : (1) s_n-\frac{1}{4}s_n = \frac{1}{4} + s_t \kern 1pc and : s_t = \displaystyle\sum_{i= 1}^{\infty} \frac{2i+1}{4^{i+1}}$
$\frac{1}{4}s_t = \frac{3}{4^{3}}+\frac{5}{4^{4}}+\frac{7}{4^{5}}+\frac{9}{4^{5}}+....$
$ s_t-\frac{1}{4} s_t = \frac{3}{4^{2}}+\frac{2}{4^{3}}+\frac{2}{4^{4}}+\frac{2}{4^{5}}$
$\frac{s_t -1\frac{1}{4}st -\frac{3}{4^{}}}{2} =\frac{1}{4^{3}}+\frac{1}{4^{4}}+\frac{1}{4^{5}}+...$
$(2) \kern 0.5pc\frac{3}{8} st - \frac{3}{32} = \displaystyle\sum_{n = 3}^{\infty} \frac{1}{4^{n}}$
$now : \displaystyle\sum_{n = 3}^{\infty} \frac{1}{4^{n}}$ is very easy to evaluate
$\displaystyle\sum_{n = 3}^{\infty} \frac{1}{4^{n}} = \frac{1}{48}$
using (2) we get:
$\frac{3}{8} st - \frac{3}{32} = \frac{1}{48}$
$\frac{3}{8} st = \frac{1}{48} + \frac{3}{32} $
$st = (\frac{1}{48} + \frac{3}{32}) \times \frac{8}{3}$
using (1) now we get:
$s_n-\frac{1}{4}s_n = \frac{1}{4} + (\frac{1}{48} + \frac{3}{32}) \times \frac{8}{3}$
$s_n = (\frac{1}{4} + (\frac{1}{48} + \frac{3}{32}) \times \frac{8}{3})\times \frac{4}{3}$
$s_n = \frac{73}{256} = 0.28515625$
there for :
$ \displaystyle\sum_{i=0} ^{ \infty} \frac{i^{2}}{4^{i}} = \frac{73}{256}$
what do you think??
