# How do you find the interval of convergence of Sigma (x+10)^n/(lnn) from n=[2,oo)?

Apr 11, 2017

The series:

${\sum}_{n = 0}^{\infty} {\left(x + 10\right)}^{n} / \ln n$

is convergent for $x \in \left[- 11 , 9\right)$ and absolutely convergent in the interior of the interval.

#### Explanation:

(i) For every $n > 2$ we have $\ln n < n$, so if $\left(x + 10\right) \ge 1$:

${\left(x + 10\right)}^{n} / \ln n > \frac{1}{n}$

and the series is divergent.

(ii) For $\left\mid x + 10 \right\mid < 1$ we have:

${\left\mid x + 10 \right\mid}^{n} / \ln n < {\left\mid x + 10 \right\mid}^{n}$

Since ${\sum}_{n = 0}^{\infty} {\left\mid x + 10 \right\mid}^{n}$ is a geometric series of ratio $r < 1$, it is convergent and then also:

${\sum}_{n = 0}^{\infty} {\left(x + 10\right)}^{n} / \ln n$ is absolutely convergent.

(iii) For $\left(x + 10\right) \le - 1$ we can write the series as an alternating series:

${\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {\left\mid x + 10 \right\mid}^{n} / \ln n$

and apply the Leibniz test:

Clearly for $\left(x + 10\right) < - 1$ we have

${\lim}_{n \to \infty} {a}_{n} = {\lim}_{n \to \infty} {\left\mid x + 10 \right\mid}^{n} / \ln n = \infty$

and the series is not convergent, while for $\left(x + 10\right) = - 1$

${\lim}_{n \to \infty} {a}_{n} = {\lim}_{n \to \infty} \frac{1}{\ln} n = 0$

and

${a}_{n + 1} / {a}_{n} = \ln \frac{n}{\ln} \left(n + 1\right) < 1$

so the series is convergent.

In conclusion the series is convergent for:

$- 1 \le x + 10 < 1$

That is for $x \in \left[- 11 , 9\right)$