Question

Determine the interval of convergence for the power series: summation n=0 to infinity (x-1)^n/ln n. Please help as I don't even know where to start!

Answer 1

ratio test for this one

Answer 2

it is pretty clearly 1 from your eyeballs, but we can take all the steps if you like

Answer 3

btw i hope the sum starts at \(n=1\) not \(n=0\) since \(\ln(0)\) is undefined

Answer 4

you start by using the ratio test
\[\vert\frac{a_{n+1}}{a_n}\vert\]

Answer 5

in this case it is \[\\\frac{(x-1)^{n+1}}{\ln(n+1)}\times \frac{\ln(n)}{(x-1)^n}\] in absolute value
took me a minute to type it
with me so far?

Answer 6

Yes. I started by using the Ratio Test and ended up with the absolute value of x-1

Answer 7

it does start at n=0 by the way

Answer 8

yes, that is right as \[\lim_{n\to \infty}\frac{\ln(n+1)}{\ln(n)}=1\]you get \[|x-1|\] as your limit
this will converge if the limit is less than \(1\) so the radius of convergence is 1, and the interval of convergence is what you get when you solve
\[|x-1|<1\] for \(x\)

Answer 9

if it starts at \(n=0\) then either it is a trick question, or more likely just a typo

Answer 10

Knowing my professor, it's a typo :)

Answer 11

lol
clear how to solve
\[|x-1|<1\] right? you do it in your head

Answer 12

not done yet however
you have to check the endpoints of the interval

Answer 13

I don't understand how to go further. I feel like the interval will be -1<x<1 but I don't understand how to go about finding and checking the endpoints.

Answer 14

hold the phone

Answer 15

the inequality you get is
\[|x-1|<1\] solve like in elementary algebra
\[|x-1|<1\\
-1<x-1<1\\
0<x<2\]

Answer 16

Lord I haven't had algebra since high school, which was 15 years ago! So this doesn't even look familiar to me

Answer 17

hey @satellite73

Answer 18

better gut used to it i guess
\(|x-1|<1\) says the distance between \(x\) and \(1\) is less than \(1\)
that is why i thought you can do it in your head

Answer 19

its demi remember you helped me with many questions thank you once again

Answer 20

oh thats ok. not right now but i will when i do

Answer 21

so we have now only to check the endpoints of the interval
i.e. replace \(x\) by \(0\) and then replace \(x\) by \(2\) and see what you get for the sum

Answer 22

with me so far?

Answer 23

if you put \(x=0\) you get \(0-1=-1\) so the sum becomes
\[\sum\frac{(-1)^n}{\ln(n)}\]

Answer 24

replace it in the inequality x-1 < 1?

Answer 25

lets back up a second to make sure it is clear how to solve these inequalities

Answer 26

If I seem a little slow on the uptake it's because I have a professor that skips from step A to step D without showing how he got there

Answer 27

thanks for dumbing it down for me

Answer 28

\[|x-a|<p\] says the distance between \(x\) and \(a\) is less than \(p\)
you solve it by starting with
\[-p<x-a<p\] then add \(a\) to both sides and get
\[-p+a<x<p+a\]

Answer 29

for example if you have to solve
\[|x-2|<3\]start with
\[-3<x-2<3\]add \(2\) and get \[-1<x<5\]

Answer 30

oh. i see now

Answer 31

if this does not ring a bell, and i am guessing it doesn't, you should review it because you will be solving lots of them

Answer 32

ok so now we got to the point where we know the limit is \(|x-1|\) and we have to make sure that is is less than 1 so we solve \[|x-1|<1\] and arrive at \(0<x<2\)

Answer 33

well my final exam is next week and there will be one question about intervals of convergence so i'm hoping i can just get throught it.....know what i mean?

Answer 34

ok. with you so far

Answer 35

yeah, but that part is not really that hard
in defence of your typo making prof, he or she probably just thought that part was obvious
now we check the endpoints of the interval
replace \(x\) by \(0\)

Answer 36

when you replace \(x\) by \(0\) you get
\[\sum\frac{(-1)^n}{\ln(n)}\] is that part clear?

Answer 37

yes

Answer 38

is it clear that because the terms alternate in sign, and the terms go to zero, that this sum converges?
that is all you have to check if it alternate
do the terms go to zero or not

Answer 39

when you say alternate in sign do you mean that it's (-1)^n when you plug in 0 and (1)^n when you plug in 2?

Answer 40

yes, that is what i mean
\((-1)^n\) is either \(-1\) or \(1\) depending on whether \(n\) is odd or even

Answer 41

ooh no no that is not what i mean
let's back up again

Answer 42

ok

Answer 43

\[\sum\frac{(-1)^n}{\ln(n)}\]is what you get when you replace \(x\) by \(0\)

Answer 44

THAT series alternates in sign because of the \((-1)^n\) part i.e. \((-1)^n\) alternates in sign
it is either \(-1\) or it is \(1\) depending, as i said, if \(n\) is odd or if \(n\) is even

Answer 45

ok

Answer 46

now if you replace \(x\) by \(2\) you get what you said
\[\sum\frac{1}{\ln(n)}\]

Answer 47

noting of course that \((1)^n=1\) no matter what \(n\) is

Answer 48

yes

Answer 49

now the question is does
\[\sum\frac{1}{\ln(n)}\]converge
what do you think?

Answer 50

Honestly, I have no idea. I really haven't understood anything we've been doing in class the past month. And convergence def confuses me.

Answer 51

\(\color{blue}{\text{Originally Posted by}}\) @satellite73
yes, that is right as \[\lim_{n\to \infty}\frac{\ln(n+1)}{\ln(n)}=1\]you get \[|x-1|\] as your limit
\(\color{blue}{\text{End of Quote}}\)

Shouldn't the limit be 0?

Answer 52

then you have some catching up to do

Answer 53

@SithsAndGiggles isn't the limit 1?

Answer 54

Oh my mistake, I read that as \(\ln(n+1)-\ln n\) for some reason... Carry on!

Answer 55

in any case the log grows very very very slowly
not fast enough for the sum to converge

Answer 56

@SithsAndGiggles same thing, same limit right?

Answer 57

Well, \(\ln(n+1)-\ln n=\ln\dfrac{n+1}{n}\), which tends to 0, but \(\dfrac{\ln(n+1)}{\ln n}\) tends to 1.

Answer 58

yes, of course

Answer 59

so no, not the same thing, not the same limit

Answer 60

so they both diverge?

Answer 61

no

Answer 62

the first one converges because it alternates and the terms go to zero
the second one is divergent because it does not alternate, and the terms do not go to zero fast enough

Answer 63

oh okay

Answer 64

so finally we see that the "interval of convergence" is
\[[0,2)\]
that took a while, hope something became clear

Answer 65

I know. I'm sorry! I really appreciate the time you spent explaining it to me.

Answer 66

you were a tremendous help

Answer 67

you might want to look and see what kind of series converge, and which series diverge
because even if there is only one "interval of convergence" question on the test
there might be a question with just series
for example
\[\sum\frac{n}{n^2+2}\]
do you think that converges?

Answer 68

oh no don't be sorry, if i hadn't wanted to help i wouldn't have
yw

Answer 69

maybe it converges based on the comparison test???

Answer 70

no it diverges
here is the easy reason
the numerator is a polynomial of degree 1
the denominator is a polynomial of degree 2
since the difference in degrees is 1 it does NOT converge
the degree of the denominator has to be larger than the degree of the numerator by MORE than 1

Answer 71

oh yes. i remember that now

Answer 72

you could compare it to the well known divergent series
\[\sum\frac{1}{n}\] if you wanted an actual test

Answer 73

if he or she is tricky, maybe a problem like
\[\sum\frac{1}{n\ln(n)}\]where the denominator is not a polynomial

Answer 74

thanks for the advise. will def look over examples

Answer 75

yw
btw integral test shows the previous one diverges as well

Answer 76

Ok. Awesome help!