Question

Find the interval of convergence of the following series (below)

Answer 1

\[\sum_{n=0}^{\infty}\frac{ nx^n }{ 4^n(n^2+1) }\]

Answer 2

Well I found the interval (-4,4), but I need to find the places that it converges absolutely and conditionally

Answer 3

And I tried using the Limit Comparison Test, and I got infinity, so I didn't know if I was allowed to move on to the Alternating Series Test

Answer 4

@jim_thompson5910

Answer 5

On the endpoints you can

Answer 6

But I tried the limit comparison test first, and I got infinity, so I did know whether that meant that the series diverges, or if it meant that I could move on to the AST

Answer 7

on the endpoint -4 , ok

Answer 8

Well not infinity, but 1, which meant that they had the same behavior, and diverged

Answer 9

can you show your work, what function did you compare it to

Answer 10

1/n

Answer 11

Well I got -4 to 4 using the ratio test

Answer 12

Then I tested -4 and I got the series \[\sum_{n=0}^{\infty}(n(-4)^n)/(4^n(n^2+1))\]

Answer 13

Which I took the absolute value of, and reduced to n/(n^2+1)

Answer 14

(-4)^n / 4^n = (-4/4)^n = (-1)^n

Answer 15

And then I compared it to 1/n

Answer 16

And I got 1

Answer 17

you can use the alternating series test for that

Answer 18

But I figured first I could just do the limit comparison test because it is first on the flowcharts that my teacher gave to us

Answer 19

Well that, the ratio test, integral test, duct

Answer 20

*dct

Answer 21

$$\Large \sum \frac{(-1)^n\cdot n }{n^2 + 1 }

$$

Answer 22

Yes

Answer 23

Then I took the absolute value of that

Answer 24

ok, then you showed that it does not converge absolutely at the point -4
but it does converge conditionally (using AST)

Answer 25

What I am really confused about is if proving that the absolute value goes to infinity shows that it is not absolutely convergent, then how do I prove that the series is divergent? I guess my original impression was that if one of those tests diverges, then the series diverges

Answer 26

$$ \Large \sum \frac{(-1)^n}{n}$$
this converges 'conditionally' , which means not taking the absolute value

Answer 27

but it diverges absolutely

Answer 28

i though that those types of tests could show divergence?

Answer 29

which tests?

Answer 30

ratio, limit comparison, direct comparison, integral

Answer 31

For example, if I were to apply the LCT to 1/(3n-1), that diverges

Answer 32

right

Answer 33

So I am confused about why in some cases it means that the series diverges, but in some cases, it just means that the series does not converge absolutely

Answer 34

sorry my browser crashed

Answer 35

no worries!

Answer 36

can you ask your question again, i will my best to help

Answer 37

with examples so i know what cases you are referring to

Answer 38

Ok thanks. I am confused about why in some cases it means that the series diverges, but in some cases, it just means that the series does not converge absolutely

Answer 39

the absolute convergence tests diverging

Answer 40

And an example is the problem that we were working on earlier

Answer 41

If that made any sense at all

Answer 42

one moment, checking again

Answer 43

Ok. Thank you!

Answer 44

so lets go through this point by point.
we found that it converges absolutely on (-4,4) , correct?

Answer 45

Yes

Answer 46

now if we plug in -4 what happens

Answer 47

Well I tried the limit comparison test

Answer 48

with 1/n, and I got 1, so they both diverge

Answer 49

we dont really use limit comparison with negative terms

Answer 50

im looking in my book, limit comparison is used only when both series have positive terms

Answer 51

Well you can test for absolute convergence can't you?

Answer 52

right, but even if it diverges absolutely you can still have conditional convergence.

Answer 53

Sorry, my wifi was going haywire

Answer 54

Ok, so then how do you prove that a series completely diverges?

Answer 55

Sorry, my wifi is on the fritz again