Question

Find the sum of the series
$\frac{1}{2} + \frac{2}{4} + \frac{3}{8} + ...$
Numerator is AP but denominator is GP how can I solve it?​

Answer 1

Solution :-

The given series is an example of A.G.P which is a combination of both arithmetic as well as geometric sequence.

Let's assume the given series be equal to S,

$\leadsto \displaystyle \sf S = \frac{1}{2} + \frac{2}{ {2}^{2} } + \frac{3}{ {2}^{3} } + ... \bigg \{ equation \: 1 \bigg\}$

Multiply both sides with 1/2

$\leadsto \displaystyle \sf \dfrac{S}2 = \frac{1}{ {2}^{2} } + \frac{2}{ {2}^{3} } + \frac{3}{ {2}^{4} } ... \bigg \{ equation \: 2\bigg\}$

Now subtracting equation 2 from equation 1

${\leadsto \displaystyle \sf S \left( 1 - \frac{1}{2} \right) = \frac{1}{2} + \frac{1}{ {2}^{2} } + \frac{1}{ {2}^{3} } + ... }$

$\leadsto \displaystyle \sf S \left( \frac{1}{2} \right) = \frac{1}{2} + \frac{1}{ {2}^{2} } + \frac{1}{ {2}^{3} } + ...$

Now, this is forming an infinite GP with first term 1/2 and common ratio 1/2. Therefore we can apply formula for sum of infinite GP.

Sum of infinite GP with first term a and common ratio r is given by,

$\pink{ \boxed {\displaystyle \sf{ S_ \infty = \frac{a}{1 - r} }}}$

By using this, we get :

$\leadsto \displaystyle \sf S \left( \frac{1}{2} \right) = \dfrac{ \frac{1}{2} }{1 - \frac{1}{2} }$

$\leadsto\displaystyle \sf S \left( \frac{1}{2} \right) = \dfrac{ \frac{1}{2} }{ \frac{1}{2} }$

$\leadsto \displaystyle \sf S \left( \frac{1}{2} \right) = 1$

Now multiplying both sides with 2, we get :

$\underline{ \boxed{ \displaystyle \sf S = 2}}$

Therefore the sum of given series is 2.