+26393 Compute 1*1/2+2*1/4+3*1/8+...+n*1/2n+...
\(\begin{array}{|rcll|} \hline S_n &=& 1*\frac{1}{2}+2*\frac{1}{4}+3*\frac{1}{8}+4*\frac{1}{16}+\ldots+n*\left(\frac{1}{2}\right)^n+\ldots \\ \hline \end{array} \)
(i)
\(\text{Multiply } S_n \text{ by } \tfrac12 \text{ the common ratio of the geometric series} \)
\(\small{ \begin{array}{rcccccccccccccl} S_n &=& 1*\frac{1}{2} &+& 2*\frac{1}{4} &+& 3*\frac{1}{8} &+& 4*\frac{1}{16} &+& \ldots &+& n*\left(\frac{1}{2}\right)^n && \\ \frac12 S_n &=& & & 1*\frac{1}{4} &+& 2*\frac{1}{8} &+& 3*\frac{1}{16} &+& \ldots &+& (n-1)*\left(\frac{1}{2}\right)^n &+& n*\left(\frac{1}{2}\right)^{n+1} \\ \hline (1-\frac12)S_n &=& [~ 1*\frac{1}{2} &+& 1*\frac{1}{4} &+& 1*\frac{1}{8} &+& 1*\frac{1}{16} &+& \ldots &+& 1*\left(\frac{1}{2}\right)^n ~] &-& n*\left(\frac{1}{2}\right)^{n+1} \quad (\text{subtract})\\ \end{array} } \)
The series in the square brackets is a geometric series with \(a = \frac12\), \(r = \frac12\) and \(n\) terms,
Thus, \(S_n\) for this series =\( \dfrac{a(1-r^{n})}{1-r} = \dfrac{\frac12(1-(\frac12)^n)}{1-\frac12}=1-(\frac12)^n\)
\(\begin{array}{rcll} (1-\frac12)S_n &=& [~ 1*\frac{1}{2} + 1*\frac{1}{4} + 1*\frac{1}{8} + 1*\frac{1}{16} + \ldots + 1*\left(\frac{1}{2}\right)^n ~] - n*\left(\frac{1}{2}\right)^{n+1} \\ (1-\frac12)S_n &=& 1-(\frac12)^n - n*\left(\frac{1}{2}\right)^{n+1} \\ \frac12S_n &=& 1-(\frac12)^n - n*\left(\frac{1}{2}\right)^{n+1} \\ \end{array} \)
(ii)
\(\text{Because } \left|\frac12\right| < 1 \text{, then } \lim \limits_{n\to \infty} { \left(\frac12 \right)^n } = 0 \text{ and } \lim \limits_{n\to \infty} { \left(\frac12 \right)^{n+1} } = 0\)
\(\begin{array}{rcll} \frac12S_n &=& 1-(\frac12)^n - n*\left(\frac{1}{2}\right)^{n+1} \\ S_n &=& 2 \left( 1-(\frac12)^n - n*\left(\frac{1}{2}\right)^{n+1} \right) \\ \lim \limits_{n\to \infty} {S_n} &=& 2 \left( 1-0 - n*0 \right) \\ &=& 2\cdot( 1 ) \\ &=& 2 \\ \end{array}\)
