Interval of Convergence for series

[erndoglai]

I'm a little stuck on a problem or i'm unsure of where to go...

The question states:
Find the opening interval of x in which the series converges

infinite
E (-1)^n (x^n)/[(n+1)2^n]
n=1

so far i've attempted to do the ratio test (An+1/An) and came up with

lim (n+1/n+2)(-x/2)
n->infinite

i got confused after that and i'm unsure but i think the next step would be the |-x/2| < 1 to show that it converges?
any help would be great!

 

[royhaas]

You are correct about the next step.

 

[soroban]

Hello, erndoglai!

One slight correction . . .

\(\displaystyle \text{Find the interval of convergence: }\;\sum^{\infty}_{n=1} \frac{(-1)^n x^n}{(n+1)2^n}\)

\(\displaystyle \text{so far i've attempted to do the ratio test }\frac{A_{n+1}}{A_n}\text{ and came up with: }\;\lim_{n\to\infty} \frac{n+1}{n+2}\left(-\frac{x}{2}\right)\)

\(\displaystyle \text{i think the next step would be: }\:\left|-\frac{x}{2}\right|\:<\<1\,\text{ to show that it converges.}\)

Basically, your work is correct, but that minus-sign can be misleading.


\(\displaystyle \text{The Ratio Test is the limit of the }absolute\; value\text{ of that fraction: }\;R \;=\;\lim_{n\to\infty}\left|\frac{A_{n+1}}{A_n}\right|\)

\(\displaystyle \text{So we can drop that }(-1)^n\text{ at the beginning: }\;\left|\frac{x^{n+1}}{(n+2)2^{n+1}}\cdot\frac{(n+1)2^n}{x^n}\right|\)

. . \(\displaystyle \text{and we have: }\;\left|\frac{2^n}{2^{n+1}}\cdot\frac{n+1}{n+2}\cdot\frac{x^{n+1}}{x^n}\right| \;=\;\left|\frac{1}{2}\cdot\frac{n+1}{n+2}\cdot x\right| \;=\;\left|\frac{n+1}{n+2}\cdot\frac{x}{2}\right|\)


\(\displaystyle \text{Then: }\;\lim_{n\to\infty}\left|\frac{n+1}{n+2}\cdot\frac{x}{2}\right| \;=\;\lim_{n\to\infty}\left|\frac{1 + \frac{1}{n}}{1 + \frac{2}{n}}\cdot\frac{x}{2}\right| \;=\;\left|\frac{1+0}{1+0}\cdot\frac{x}{2}\right| \;=\;\left|\frac{x}{2}\right|\)



\(\displaystyle \text{Moral: When performing the Ratio Test on an alternating series,}\)
. . . . . . . . . . . . . . . \(\displaystyle \text{drop the "alternator".}\)

 

[BigGlenntheHeavy]

\(\displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n}x^{n}}{(n+1)2^{n}}. \ Now, \ to \ test \ for \ convergence, \ apply \ the \ ratio \ test.\)

\(\displaystyle Ergo, \ \lim_{n\to\infty} \ \bigg|\frac{x^{n+1}}{(n+2)2^{n+1}}*\frac{(n+1)2^{n}}{x^{n}}\bigg|, \ Note: \ |(-1)^{?}| \ = \ 1.\)

\(\displaystyle = \ \lim_{n\to\infty} \ \bigg|\frac{x(n+1)}{2(n+2)}\bigg| \ = \ \frac{|x|}{2} \ \lim_{n\to\infty}\bigg|\frac{n+1}{n+2}\bigg|\)

\(\displaystyle = \ \frac{|x|}{2}(1) \ = \ \frac{|x|}{2} \ < \ 1 \ (ratio \ test).\)

\(\displaystyle = \ |x| \ < \ 2 \ \implies \ -2< \ x \ < \ 2, \ therefore \ converges \ on \ (-2,2), \ but \ what \ about \ the \ endpoints?\)

\(\displaystyle When \ x \ = \ -2, \ we \ have \ the \ divergent \ harmonic \ series \ and \ when \ x \ = \ 2, \ we \ have \ convergent\)

\(\displaystyle \ alternating \ series.\)

\(\displaystyle Hence, \ series \ converges \ on \ (-2, 2 \ ]. \ QED\)

 

[Deleted member 4993]

Glen,

Nice work - but you did not use Q.E.D.!!