For What values of x does the series (x-3)^n/x converge?

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Suresh Chand · Accepted Answer

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_{nrightarrow ∞} 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 enspaceenspaceenspace i.e. enspaceenspace -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

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

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

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