Question

Find the radius and the interval of convergence

sigma (X^k)/(k+1)

Answer 1

first take the ratio, and work the limit

Answer 2

\[\sum_{k=0}^{\infty} \frac{ X^k }{ k+1 }\]

Answer 3

I did that well. I am having troubles testing the limits

Answer 4

hmm, show your work, lets see how well you got to that point then

Answer 5

I dont understand how to test if it decreasing in the alternating series test

Answer 6

so your trying to test for x=-1 ?

Answer 7

give me a second

Answer 8

\[\lim_{k\to\infty}\frac{x^{n+1}/(k+2)}{x^n/(k+1)}\]
\[\lim_{k\to\infty}x\frac{k+1}{k+2}\]
\[|x|\lim_{k\to\infty}\frac{k+1}{k+2}=|x|\]

Answer 9

when x=1, this is a harmonic series ...

when x=-1, its an alternating harmonic series ...

do you recall the convergence of those?

Answer 10

harmonic no

Answer 11

how do you do it?

Answer 12

the harmonic does not converge, the alternating does ...

Answer 13

but why? Could you explain?

Answer 14

what is a harmonic series? and how do i know if it converges or diverges?

Answer 15

harmonic is\[\frac11+\frac12+\frac13+\frac14+\frac15+...\]
the proofing is textbook, but it eludes me

Answer 16

so is like \[\frac{ 1 }{ K! }\]

Answer 17

http://web.williams.edu/Mathematics/lg5/harmonic.pdf

Answer 18

ohh my bad \[\frac{ 1 }{ k }\]

Answer 19

ohh ok i see if it has a K on the bottom

Answer 20

@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.

Answer 21

I still dont get how can someone recognize that a function is a harmonic series just by looking at it

Answer 22

the alternating harmonic series converges :) that proof eludes me as well ... just recall it from the classes.

Answer 23

is because of the K?

Answer 24

1/n is the harmonic series by definition

Answer 25

yes

Answer 26

so K by itself would be a harmonic series?

Answer 27

2K +1

Answer 28

1/k^p converges for p > 1

Answer 29

10 k -2

Answer 30

isnt it 1/k^p a p-series?

Answer 31

yes

Answer 32

ok sooo if it has a K is a harmonic series and if it has an exponent is a p series?

Answer 33

1/k = 1/k^1

by definition, 1/k is the harmonic series.

Answer 34

that is why the p series test says that if p > 1 it converges, but if 0<p<= 1 it diverges

Answer 35

so it diverges?

Answer 36

why do i feel like this is just going in circles ....

Answer 37

and please special help in the alternating series one !

Answer 38

sorry i am just reinforcing the knowledge you are giving me by enphasing what you say lol

Answer 39

the alternating harmonic series converges, by properties that i cant recall either. they are explained better by the texts

Answer 40

http://www.math.com/tables/expansion/tests.htm

Answer 41

radius = 1

interval is: [-1,1)

Answer 42

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

Answer 43

im getting tired as well, but there are plenty of smarter people then me about :)