blackstreet23
  • blackstreet23
Find the radius and the interval of convergence sigma (X^k)/(k+1)
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
amistre64
  • amistre64
first take the ratio, and work the limit
blackstreet23
  • blackstreet23
\[\sum_{k=0}^{\infty} \frac{ X^k }{ k+1 }\]
blackstreet23
  • blackstreet23
I did that well. I am having troubles testing the limits

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

amistre64
  • amistre64
hmm, show your work, lets see how well you got to that point then
blackstreet23
  • blackstreet23
I dont understand how to test if it decreasing in the alternating series test
amistre64
  • amistre64
so your trying to test for x=-1 ?
blackstreet23
  • blackstreet23
give me a second
amistre64
  • amistre64
\[\lim_{k\to\infty}\frac{x^{n+1}/(k+2)}{x^n/(k+1)}\] \[\lim_{k\to\infty}x\frac{k+1}{k+2}\] \[|x|\lim_{k\to\infty}\frac{k+1}{k+2}=|x|\]
blackstreet23
  • blackstreet23
1 Attachment
amistre64
  • amistre64
when x=1, this is a harmonic series ... when x=-1, its an alternating harmonic series ... do you recall the convergence of those?
blackstreet23
  • blackstreet23
harmonic no
blackstreet23
  • blackstreet23
how do you do it?
amistre64
  • amistre64
the harmonic does not converge, the alternating does ...
blackstreet23
  • blackstreet23
but why? Could you explain?
blackstreet23
  • blackstreet23
what is a harmonic series? and how do i know if it converges or diverges?
amistre64
  • amistre64
harmonic is\[\frac11+\frac12+\frac13+\frac14+\frac15+...\] the proofing is textbook, but it eludes me
blackstreet23
  • blackstreet23
so is like \[\frac{ 1 }{ K! }\]
amistre64
  • amistre64
http://web.williams.edu/Mathematics/lg5/harmonic.pdf
blackstreet23
  • blackstreet23
ohh my bad \[\frac{ 1 }{ k }\]
blackstreet23
  • blackstreet23
ohh ok i see if it has a K on the bottom
beginnersmind
  • beginnersmind
@amistre "the proofing is textbook, but it eludes me" 1/3 + 1/4 > 1/2 1/5 + 1/6 + 1/7 + 1/8 > 1/2 1/9 + ... 1/16 > 1/2 etc.
blackstreet23
  • blackstreet23
I still dont get how can someone recognize that a function is a harmonic series just by looking at it
amistre64
  • amistre64
the alternating harmonic series converges :) that proof eludes me as well ... just recall it from the classes.
blackstreet23
  • blackstreet23
is because of the K?
amistre64
  • amistre64
1/n is the harmonic series by definition
amistre64
  • amistre64
yes
blackstreet23
  • blackstreet23
so K by itself would be a harmonic series?
blackstreet23
  • blackstreet23
2K +1
amistre64
  • amistre64
1/k^p converges for p > 1
blackstreet23
  • blackstreet23
10 k -2
blackstreet23
  • blackstreet23
isnt it 1/k^p a p-series?
amistre64
  • amistre64
yes
blackstreet23
  • blackstreet23
ok sooo if it has a K is a harmonic series and if it has an exponent is a p series?
amistre64
  • amistre64
1/k = 1/k^1 by definition, 1/k is the harmonic series.
amistre64
  • amistre64
that is why the p series test says that if p > 1 it converges, but if 0
blackstreet23
  • blackstreet23
so it diverges?
amistre64
  • amistre64
why do i feel like this is just going in circles ....
blackstreet23
  • blackstreet23
and please special help in the alternating series one !
blackstreet23
  • blackstreet23
sorry i am just reinforcing the knowledge you are giving me by enphasing what you say lol
amistre64
  • amistre64
the alternating harmonic series converges, by properties that i cant recall either. they are explained better by the texts
amistre64
  • amistre64
http://www.math.com/tables/expansion/tests.htm
amistre64
  • amistre64
radius = 1 interval is: [-1,1)
blackstreet23
  • blackstreet23
Ohh ok. I am so sleepy right now haha. I will analyse it more tomorrow :). Thanks a lot ! anything I will send you more messages :D
amistre64
  • amistre64
im getting tired as well, but there are plenty of smarter people then me about :)

Looking for something else?

Not the answer you are looking for? Search for more explanations.