# Math Help - finding the interval of convergence..

1. ## finding the interval of convergence..

in the first problem i was asked to find the interval of convergence and then i was supposed to check the endpoints, im not sure if i did this problem right and was wondering if someone could check my work and tell me where i went wrong. in the next problem i was supposed to find a geometric series, which seemed easy enough, but please check both problems. thanks in advance..IMG.pdf

2. Originally Posted by slapmaxwell1
in the first problem i was asked to find the interval of convergence and then i was supposed to check the endpoints, im not sure if i did this problem right and was wondering if someone could check my work and tell me where i went wrong. in the next problem i was supposed to find a geometric series, which seemed easy enough, but please check both problems. thanks in advance..IMG.pdf

The first series is one around $x=2$ and its convergence radius is $\frac{1}{3}$ , since you got $3|x-2|<1\iff |x-2|<\frac{1}{3}$ .

Besides this, the series converges at $x=\frac{5}{3}$ (you get the alternating harmonic series), but diverges at $\frac{7}{3}$ (you get the harmonic series), so the interval of conv. is $\left[\left. \frac{5}{3}, \frac{7}{3}\right.\right)$.

The second question is written, apparenty, in sumerian mixed with ancient sanscrit, but if I have to guess you're asked to find a power series for that function h(x), and yes: what you did is correct.

Tonio

3. Fortunately, I speak both Sumerian and Sanscrit (as well as practice reading really bad handwriting- mine)! The first line is "Find a geometric power series centered at 0".

4. i was really tired when i did those problems, besides prep for the calc final i had to write an economic analysis?? thanks again im going to definitely start doing more problems when im not so tired. LOL

5. Originally Posted by HallsofIvy
Fortunately, I speak both Sumerian and Sanscrit (as well as practice reading really bad handwriting- mine)! The first line is "Find a geometric power series centered at 0".

Fine , but if the series was to be centered at zero then it perhaps is nicer to write it as

$\sum^\infty_{n=1}\frac{3}{2}\left(-\frac{x}{2}\right)^n=\sum^\infty_{n=0}(-1)^n\frac{3}{2^{n+1}}\,x^n$

Tonio