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

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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

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