# For What values of x does the series  $$\sum_{n=1}^{∞} \frac{(x – 3)^n}{x}$$ converge?

Solution:

Given series is

$$\sum_{n=1}^{∞} \frac{(x – 3)^n}{x}$$          …. (i)

comparing it with $$\sum_{n=1}^{∞} u_n$$ then

$$u_n = \frac{(x – 3)^n}{x}$$         ….. (ii)

Here

$$\lim_{n\rightarrow ∞} \left [ \frac{(x – 3)^{n+1}}{x} \times \frac{x}{(x – 3)^n} \right ]$$

$$= \lim_{ n \to ∞} (x – 3)$$

= x – 3

By D’Alembert ratio test the given series is convergent for

$$\left | x – 3 \right | < 3 \enspace\enspace\enspace i.e. \enspace\enspace -1 < (x – 3) < 1$$

= 2 < x < 4

and is divergent for $$\left | x – 3 \right | > 1$$

And further test is needed for $$\left | x – 3 \right | = 1$$  i.e. at x = 2 and x = 4

At x = 2, the series (i) becomes,

$$\sum_{n=1}^{∞}(u_n) = \sum_{n=1}^{∞} \frac{(-1)^n}{2}$$

And is an alternative series. Comparing it with $$\sum_{n=1}^{∞} (-1)^n \enspace V_n$$ then

$$V_n = \frac{1}{2}$$

Here

$$\lim_{n \to ∞} V_n = \lim_{b \to ∞}\left ( \frac{1}{2} \right ) = \frac{1}{2} \neq 0$$

So, the given series is divergent at x = 2, by Leibnitz test.

And at x = 4, the series (i) becomes

$$= \sum_{n=1}^{∞} (u_n)$$

$$= \sum_{n=1}^{∞} \left ( \frac{1}{2} \right )$$

$$= \left ( \frac{1}{2} \right ) \sum_{n=1}^{∞} (1)$$

= ∞

This means the given series is divergent at x = 4

Thus the given series is convergent only for 2 < x < 4