What is the sum of the first n terms of the series 1/2+3/4+5/8+... ?

Feb 15, 2018

${S}_{n} = 3 - \frac{2 n + 3}{2} ^ n$

Explanation:

Let ${S}_{n} = \frac{1}{2} + \frac{3}{4} + \frac{5}{8} + \ldots \ldots \ldots \ldots \ldots \ldots \text{n terms}$

What is ${n}^{t h}$ term of the series? Well numerators are in arithmatic sequence and numerator of ${n}^{t h}$ term is $2 n - 1$ and denominator is in geometric series with ${n}^{t h}$ term being ${2}^{n}$. Hence ${n}^{t h}$ term of given series is $\frac{2 n - 1}{2} ^ n$ and series is

${S}_{n} = \frac{1}{2} + \frac{3}{4} + \frac{5}{8} + \frac{7}{16} + \ldots \ldots \ldots \ldots \ldots \ldots + \frac{2 n - 1}{2} ^ n$ and then

$\frac{1}{2} {S}_{n} = \frac{1}{4} + \frac{3}{8} + \frac{5}{16} + \ldots \ldots \ldots \ldots \ldots \ldots + \frac{2 n - 1}{2} ^ \left(n + 1\right)$

Subtracting latter from former, we get

${S}_{n} - \frac{1}{2} {S}_{n} = \frac{1}{2} + \frac{2}{4} + \frac{2}{8} + \ldots . + \frac{2}{2} ^ n - \frac{2 n - 1}{2} ^ \left(n + 1\right)$

or $\frac{1}{2} {S}_{n} = \frac{1}{2} + \frac{2}{4} \left[1 + \frac{1}{2} + \ldots . + \frac{1}{2} ^ \left(n - 2\right)\right] - \frac{2 n - 1}{2} ^ \left(n + 1\right)$

= $\frac{1}{2} + \frac{1}{2} \frac{1 - \frac{1}{2} ^ \left(n - 1\right)}{1 - \frac{1}{2}} - \frac{2 n - 1}{2} ^ \left(n + 1\right)$

= $\frac{1}{2} + 1 - \frac{1}{2} ^ \left(n - 1\right) - \frac{2 n - 1}{2} ^ \left(n + 1\right)$

= $\frac{3}{2} - \frac{4 + 2 n - 1}{2} ^ \left(n + 1\right)$

= $\frac{3}{2} - \frac{2 n + 3}{2} ^ \left(n + 1\right)$

and ${S}_{n} = 3 - \frac{2 n + 3}{2} ^ n$