Find the interval of convergence for $\sum_{n = 0}^{\infty} \frac{n x^{n}}{5^{n}}$

By the ratio test, one finds the radius of convergence. That is,

p = lim_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} | = lim_{n \to \infty} | \frac{(n + 1) x^{n + 1}}{5^{n + 1}} • \frac{5^{n}}{n x^{n}} | = lim_{n \to \infty} \frac{(n + 1) x^{n} . x {.5}^{n}}{5^{n} .5 . n . x^{n}} | = \frac{| x |}{5}

Therefore, the series converges if

\frac{| x |}{5} < 1

| x | < 5

. Hence the series is convergent if

| x | < 5

Now one has to check separately the endpoints

x = - 5

and

x = 5

. In this case, the series diverges at

x = \pm 5

. Because for

x = 5

, the series is

\sum_{n = 0}^{\infty} n = \infty

(the general term does not go to zero). Also for

x = - 5

, the series is

\sum_{n = 0}^{\infty} {(- 1)}^{n} n

, which also diverges since the general term does not go to zero.

By the ratio test, one finds the radius of convergence. That is,

p = lim_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} | = lim_{n \to \infty} | \frac{(n + 1) x^{n + 1}}{5^{n + 1}} • \frac{5^{n}}{n x^{n}} | = lim_{n \to \infty} \frac{(n + 1) x^{n} . x {.5}^{n}}{5^{n} .5 . n . x^{n}} | = \frac{| x |}{5}

Therefore, the series converges if

\frac{| x |}{5} < 1

| x | < 5

. Hence the series is convergent if

| x | < 5

Now one has to check separately the endpoints

x = - 5

and

x = 5

. In this case, the series diverges at

x = \pm 5

. Because for

x = 5

, the series is

\sum_{n = 0}^{\infty} n = \infty

(the general term does not go to zero). Also for

x = - 5

, the series is

\sum_{n = 0}^{\infty} {(- 1)}^{n} n

, which also diverges since the general term does not go to zero.