Find the values of x for which the series converges (x + 3)ⁿ/2ⁿ

Solution :

The series (x + 3)ⁿ/2ⁿ can expanded as below:

\(\sum_{n=1}^{n}\frac{(x+3)^{n}}{2^{n}}\) = \(\sum_{n=1}^{n}\frac{(x+3)^{n}}{2^{n}} = \frac{(x+3)}{2} + \frac{(x+3)^{2}}{2^{2}} + \frac{(x+3)^{3}}{2^{3}}+ .....+ \frac{(x+3)^{n}}{2^{n}} + ......\)

The common ratio r of the above series is (x + 3)/2.

For series to converge the following inequality has to be satisfied:

0 < |(x + 3)/2| < 1

So looking at the left Hand side inequality we have:

0 < (x + 3)/2 ⇒ x > -3

Then considering the right hand side inequality we have:

(x + 3)/2 < 1

⇒ x < 2 - 3

x < -1

Combining the two inequalities we finally get

-3 < x < -1

Hence the series will converge for the values of x as stipulated by the inequality -3 < x < -1.

So when r < 0,

Sum of the series will be as follows

Sum of the series = 1/(1 - r)

= 1/(1 - (x + 3)/2)

= 1/((2 - x - 3)/2

= 2/(-1 - x)

= -2/(1 + x)

So we can finally state that:

-2/(1 + x) = \(\sum_{n=1}^{n}\frac{(x+3)^{n}}{2^{n}} = \frac{(x+3)}{2} + \frac{(x+3)^{2}}{2^{2}} + \frac{(x+3)^{3}}{2^{3}}+ .....+ \frac{(x+3)^{n}}{2^{n}} + ......\)

When -3 < x < -1