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blackstreet23
 one year ago
Find the radius and the interval of convergence
sigma (X^k)/(k+1)
blackstreet23
 one year ago
Find the radius and the interval of convergence sigma (X^k)/(k+1)

This Question is Closed

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1first take the ratio, and work the limit

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0\[\sum_{k=0}^{\infty} \frac{ X^k }{ k+1 }\]

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0I did that well. I am having troubles testing the limits

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1hmm, show your work, lets see how well you got to that point then

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0I dont understand how to test if it decreasing in the alternating series test

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1so your trying to test for x=1 ?

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0give me a second

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1\[\lim_{k\to\infty}\frac{x^{n+1}/(k+2)}{x^n/(k+1)}\] \[\lim_{k\to\infty}x\frac{k+1}{k+2}\] \[x\lim_{k\to\infty}\frac{k+1}{k+2}=x\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1when x=1, this is a harmonic series ... when x=1, its an alternating harmonic series ... do you recall the convergence of those?

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0how do you do it?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1the harmonic does not converge, the alternating does ...

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0but why? Could you explain?

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0what is a harmonic series? and how do i know if it converges or diverges?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1harmonic is\[\frac11+\frac12+\frac13+\frac14+\frac15+...\] the proofing is textbook, but it eludes me

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0so is like \[\frac{ 1 }{ K! }\]

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0ohh my bad \[\frac{ 1 }{ k }\]

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0ohh ok i see if it has a K on the bottom

beginnersmind
 one year ago
Best ResponseYou've already chosen the best response.0@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
 one year ago
Best ResponseYou've already chosen the best response.0I still dont get how can someone recognize that a function is a harmonic series just by looking at it

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1the alternating harmonic series converges :) that proof eludes me as well ... just recall it from the classes.

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0is because of the K?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.11/n is the harmonic series by definition

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0so K by itself would be a harmonic series?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.11/k^p converges for p > 1

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0isnt it 1/k^p a pseries?

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0ok sooo if it has a K is a harmonic series and if it has an exponent is a p series?

amistre64
 one year ago
Best ResponseYou've already chosen the best response.11/k = 1/k^1 by definition, 1/k is the harmonic series.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1that is why the p series test says that if p > 1 it converges, but if 0<p<= 1 it diverges

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1why do i feel like this is just going in circles ....

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0and please special help in the alternating series one !

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0sorry i am just reinforcing the knowledge you are giving me by enphasing what you say lol

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1the alternating harmonic series converges, by properties that i cant recall either. they are explained better by the texts

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1radius = 1 interval is: [1,1)

blackstreet23
 one year ago
Best ResponseYou've already chosen the best response.0Ohh 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
 one year ago
Best ResponseYou've already chosen the best response.1im getting tired as well, but there are plenty of smarter people then me about :)
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