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# Compute 1*1/2+2*1/4+3*1/8+...+n*1/2n+...

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Compute 1*1/2+2*1/4+3*1/8+...+n*1/2n+...

Sep 20, 2017

#1
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∑[(n*1/2^n), n, 1, ∞] =converges to 2.

Sep 20, 2017
#2
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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)

$$\text{Because } \left|\frac12\right| < 1 \text{, then } \lim \limits_{n\to \infty} { \left(\frac12 \right)^n } = 0 \text{ and } \lim \limits_{n\to \infty} { \left(\frac12 \right)^{n+1} } = 0$$

$$\begin{array}{rcll} \frac12S_n &=& 1-(\frac12)^n - n*\left(\frac{1}{2}\right)^{n+1} \\ S_n &=& 2 \left( 1-(\frac12)^n - n*\left(\frac{1}{2}\right)^{n+1} \right) \\ \lim \limits_{n\to \infty} {S_n} &=& 2 \left( 1-0 - n*0 \right) \\ &=& 2\cdot( 1 ) \\ &=& 2 \\ \end{array}$$

Sep 20, 2017