# \sum_(n=2)^\infty ((x+2)^n)/(2^n\lnn)?

## "Find the radius of convergence and interval of convergence." 11.8 ${a}_{n} = \frac{{\left(x + 2\right)}^{n}}{{2}^{n} \setminus \ln n}$ ${a}_{n + 1} = \frac{{\left(x + 2\right)}^{n + 1}}{{2}^{n + 1} \setminus \ln \left(n + 1\right)} = \frac{{\left(x + 2\right)}^{n} \left(x + 2\right)}{{2}^{n} \left(2\right) \setminus \ln \left(n + 1\right)}$ Applying Ratio Test: $\setminus {\lim}_{n \setminus \rightarrow \setminus \infty} | {a}_{n + 1} / {a}_{n} | = \setminus \stackrel{L}{\setminus \infty} | \frac{\left(x + 2\right) \setminus \ln n}{2 \setminus \ln \left(n + 1\right)} | \ldots$ What should I do next? I know that after I find the ROC with $\setminus \left\mid x \right\mid \setminus < 1$, I should test the endpoints... ...but I can't even get those right now. I'm stuck on the Ratio Test.

Apr 23, 2018

$R = 2 ,$ interval of convergence: $\left[- 4 , 0\right)$

#### Explanation:

So, lifting off from where you ended, let's apply the Ratio Test:

${a}_{n + 1} = {\left(x + 2\right)}^{n + 1} / \left({2}^{n + 1} \ln \left(n + 1\right)\right)$
${a}_{n} = {\left(x + 2\right)}^{n} / \left({2}^{n} \ln n\right)$

Thus,

$L = {\lim}_{n \to \infty} | {\left(x + 2\right)}^{n + 1} / \left({2}^{n + 1} \ln \left(n + 1\right)\right) \cdot \frac{{2}^{n} \ln n}{x + 2} ^ n |$

${\left(x + 2\right)}^{n + 1} / {\left(x + 2\right)}^{n} = \left(x + 2\right)$

${2}^{n} / {2}^{n + 1} = \frac{1}{2}$

Thus, factoring the $\frac{1}{2}$ and $\left(x + 2\right)$ outside, maintaining its absolute value, we get

$\frac{1}{2} | x + 2 | {\lim}_{n \to \infty} \ln \frac{n}{\ln} \left(n + 1\right)$ (We drop the absolute values on the logarithms as we're heading toward infinity -- everything is positive)

${\lim}_{n \to \infty} \ln \frac{n}{\ln} \left(n + 1\right) = \frac{\infty}{\infty}$ -- indeterminate, we should quickly apply l'Hospital's Rule, rewriting with a hypothetical differentiable variable $y$, as $n$ is not differentiable:

${\lim}_{n \to \infty} \ln \frac{n}{\ln} \left(n + 1\right) = {\lim}_{y \to \infty} \ln \frac{y}{\ln} \left(y + 1\right) = {\lim}_{y \to \infty} \frac{\frac{1}{y}}{\frac{1}{y + 1}} = {\lim}_{y \to \infty} \frac{y + 1}{y} = 1$

Thus, we see ${\lim}_{n \to \infty} \ln \frac{n}{\ln} \left(n + 1\right) = 1 ,$ and we know we have convergence when $L < 1$ or

$\frac{1}{2} | x + 2 | < 1$

$| x + 2 | < 2 \to R = 2$

Now, determine the interval:

$- 2 < x + 2 < 2$

$- 4 < x < 0$

Test these endpoints:

$x = 0 :$

${\sum}_{n = 2}^{\infty} \frac{\cancel{{2}^{n}}}{\cancel{{2}^{n}} \ln n} = {\sum}_{n = 2}^{\infty} \frac{1}{\ln} n$

We'll use the Direct Comparison Test: $\frac{1}{n} \le \frac{1}{\ln} \left(n\right)$ on $\left[2 , \infty\right)$, we know this because the logarithm grows slower, and the smaller denominator ensures a larger sequence, so, since the smaller series ${\sum}_{n = 2}^{\infty} \frac{1}{n}$ diverges by the $p -$series test where $p = 1$, so does the larger series.

This endpoint is thus excluded from the interval of convergence.

$x = - 4 :$

${\sum}_{n = 2}^{\infty} {\left(- 2\right)}^{n} / \left({2}^{n} \ln n\right) = {\sum}_{n = 2}^{\infty} \frac{{\left(- 1\right)}^{n} \cancel{{2}^{n}}}{\cancel{{2}^{n}} \ln n} = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} / \ln n$

Using the Alternating Series Test, we see ${b}_{n} = \frac{1}{\ln} n$ is decreasing on $\left[2 , \infty\right)$ due to the growing denominator and ${\lim}_{n \to \infty} \frac{1}{\ln} n = 0$

So, for this endpoint, we have convergence by the Alternating Series Test.

Interval of Convergence: $\left[- 4 , 0\right)$

Apr 23, 2018

The interval of convergence is $- 4 \le x < 0$

#### Explanation:

${\lim}_{n \to + \infty} | {a}_{n + 1} / {a}_{n} | = {\lim}_{n \to + \infty} | \frac{{\left(x + 2\right)}^{n + 1} / \left({2}^{n + 1} \ln \left(n + 1\right)\right)}{{\left(x + 2\right)}^{n} / \left({2}^{n} \ln \left(n\right)\right)} |$

$= {\lim}_{n \to + \infty} | \frac{x + 2}{2} \ln \frac{n}{\ln \left(n + 1\right)} |$

$= | \frac{x + 2}{2} |$ as ${\lim}_{n \to + \infty} | \ln \frac{n}{\ln \left(n + 1\right)} | = 1$

The series converges for

$| \frac{x + 2}{2} | < 1$

$- 2 < x + 2 < 2$

$- 4 < x < 0$

For $x = 0$

${\sum}_{n = 2}^{\infty} {\left(x + 2\right)}^{n} / \left({2}^{n} \ln \left(n\right)\right) = {\sum}_{n = 2}^{\infty} {2}^{n} / \left({2}^{2} \ln n\right) = {\sum}_{n = 2}^{\infty} \frac{1}{\ln} n$

The series diverges for $x = 0$

For $x = - 4$

${\sum}_{n = 2}^{\infty} {\left(- 4 + 2\right)}^{n} / \left({2}^{n} \ln \left(n\right)\right) = {\sum}_{n = 2}^{\infty} {\left(- 2\right)}^{n} / \left({2}^{2} \ln n\right) = {\sum}_{n = 2}^{\infty} {\left(- 1\right)}^{n} / \left(\ln n\right)$

This series converges conditionally for $x = - 4$

The interval of convergence is $- 4 \le x < 0$