# Find the radius of convergence and the interval of conv...

• Jul 18th 2013, 06:33 PM
hbarnes
Find the radius of convergence and the interval of conv...
convergence, also indicating the type of convergence, either absolutely or conditionally.

Σ [n=2 to n=infinity] [(-1)^n (2x+3)^n]/(nln(n))

I tried to take the absolute root test to cancel the nth powers, but I am stumped by the natural log. Can someone please tell me where to start?

• Jul 18th 2013, 07:07 PM
Prove It
Re: Find the radius of convergence and the interval of conv...
Have you tried the ratio test?
• Jul 18th 2013, 08:18 PM
Soroban
Re: Find the radius of convergence and the interval of conv...
Hello, hbarnes!

Quote:

Find the interval of converence.
Also, indicate the type of convergence: absolute or conditional.

. . $\displaystyle \displaystyle \sum^{\infty}_{n=2} \frac{(-1)^n(2x+3)^n}{n\ln(n)}$

I tried to take the absolute root test to cancel the nth powers,
but I am stumped by the natural log.

Ratio Test

$\displaystyle R \:=\:\left|\frac{a_{n+1}}{a_n}\right| \:=\:\left|\frac{(2x+3)^{n+1}}{(n+1)\ln(n+1)}\cdot \frac{n\ln(n)}{(2x+3)^n}\right|$

. . . . . . . . . $\displaystyle =\;\left|\frac{(2x+3)^{n+1}}{(2x+3)^n}\cdot\frac{n }{n+1}\cdot\frac{\ln(n)}{\ln(n+1)}\right|$

. . . . . . . . . $\displaystyle =\;\left|(2x+3)\cdot\frac{\frac{1}{n}}{\frac{1}{n+ 1}} \cdot\frac{\ln(n)}{\ln(n+1)}\right|$

$\displaystyle \displaystyle\lim_{n\to\infty}R \;=\;\lim_{n\to\infty}|2x+3|\cdot\lim_{n\to\infty} \left|\frac{\frac{1}{n}}{\frac{1}{n+1}}\right| \cdot \lim_{n\to\infty}\left|\frac{\ln(n)}{\ln(n+1)} \right|$

$\displaystyle \displaystyle\lim_{n\to\infty}R \;=\;|2x+3|\cdot 1 \cdot \lim_{n\to\infty}\left|\frac{\ln(n)}{\ln(n+1)} \right|$ . [1]

For that last limit, apply L'Hopital:

. . $\displaystyle \lim_{n\to\infty}\frac{\frac{1}{n}}{\frac{1}{n+1}} \;=\;\lim_{n\to\infty}\frac{n+1}{n} \;=\;\lim_{n\to\infty}\frac{1+\frac{1}{n}}{1} \;=\;1$

Hence, [1] becomes: .$\displaystyle |2x+3|\cdot 1\cdot 1 \:=\:|2x+3|$

We have:

. . $\displaystyle \begin{array}{c}\qquad\;\;\;|2x+3| \:<\:1 \\ \\ -1 \:<\:2x+3\:<\:1 \\ \\ -4\:<\:2x\:<\:-2 \\ \\ -2\:<\:x\:<\:-1 \end{array}$

The series is absolutely convergent.
• Jul 18th 2013, 08:33 PM
Prove It
Re: Find the radius of convergence and the interval of conv...
That may not be the complete interval of convergence, as we don't know anything about the convergence of the series when the ratio of terms is 1.

So substitute x = -2 and x = -1 into your function and use a different test to determine the convergence of those two series.