What ? Sequence , Ah Ok!

These are two geometric progressions written in inter wined form.

$1^{st}$ is $1,\frac{-1}{3}, \frac{1}{9} , \frac{-1}{27} .....$ this is a GP with $a=1$ and $r= \frac{-1}{3}$

$2^{nd}$ GP is $\frac{1}{2},\frac{1}{4},\frac{1}{8},\frac{1}{16} , ....$ this is a GP with $a=\frac{1}{2}$ and $r=\frac{1}{2}$

Sum of infinite terms of a GP is $\dfrac{a}{1-r}$

Thus the sum of infinite terms of 1st GP is $\frac{1}{1-(\frac{-1}{3})} = \frac{1}{\frac{4}{3}} = \dfrac{3}{4}$

The sum of infinite terms of 2nd GP is $\dfrac{\frac{1}{2}}{1-\frac{1}{2}}= \dfrac{\frac{1}{2}}{\frac{1}{2}}=1$

Hence sum of the whole sequence infinite terms is $1 + \frac{3}{4} = \dfrac{7}{4}$

Thus the asked number $a+b$ is $7+4 = \boxed{11}$

Adding up the two infinite G.P's will give you the answer..

$\frac{-1}{3}, \frac{1}{9}......$ and $\frac{1}{2}, \frac{1}{4}.......$

$\rightarrow S=\frac{1}{1-\frac{-1}{3}}+\frac{\frac{1}{2}}{1-\frac{1}{2}}$ ........................Sum of an infinite GP with $r<1$ $=\frac{a}{1-r}$

$\rightarrow S=\frac{3}{4}+1$

$\rightarrow \boxed {S=\frac{7}{4}}$

So, $a+b=\boxed {11}$