What is the sum of first eight terms of the series \(1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+...?\)

Concept:

Sum of Geometric progression:

For a G.P. with n terms, common ratio r and first term a, the sum of n terms is given by:

1. \(\rm S_n=\dfrac{a(r^n-1)}{r-1}\ \ \ \ \{...r>1\)
2. \(\rm S_n=\dfrac{a(1-r^n)}{1-r}\ \ \ \ \{...r<1\)

Calculation:

Notice that the given summation is a G.P with common ratio \(\rm r = -\dfrac{1}{2}\).

Since r < 1, the sum of the G.P is given by the formula 2.

For the given G.P. \(\rm a=1, r=-\dfrac{1}{2}\mbox{ and } n = 8\).

Thus, the required sum is given as follows:

\(\rm \begin{align*} S_8 &= \dfrac{a(1-r^n)}{1-r}\\ &= \dfrac{1\left(1-\left(-\frac{1}{2}\right)^8\right)}{1+\frac{1}{2}}\\ &= \dfrac{1\left(1-\left(\frac{1}{256}\right)\right)}{\frac{3}{2}}\\ &= \dfrac{2\left(\frac{255}{256}\right)}{3}\\ &= \dfrac{2\times 255}{3\times 256}\\ &= \dfrac{85}{128} \end{align*}\)

Therefore, the sum of the first 8 terms of the series \(1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\cdots\) is \(\dfrac{85}{128}\).