Finding Interval of Convergence for Alternating Series

[Rasine]

am trying to find the intervals of convergence for the summation, first deritive, and intergral of problems like this:

the sum of [(-1)^n+1(x-5)^n]/[n5^n] from n=1 to infinity

i know it is an alternating series and thus i am attempting to use that test to find convergence/diverigence
lim as n gose to infinty=0 and asub.n is less than or equal to a

when i try to find the limit, both the top and bottom go to infinity and then I'm trying to use l'hopital rule for an indeterminat form, but that's not wokring too well

maybe i need to use the raito test, but that doesn't seem to work well either

please point me in the right direction

 

[d_leet]

The ratio test should work perfectly for this series.

 

[Rasine]

when i do the raito test...the lim as n goes to infinty is not 0...the series diverges ... how dose that work out...am i wrong?

 

[d_leet]

when i do the raito test...the lim as n goes to infinty is not 0...the series diverges ... how dose that work out...am i wrong?

I think you're wrong... Can you show your work using the ratio test?

 

[Rasine]

well, for the raito test one condition has to be met inorder for the series to converge, that is the abs(an-10/an)<1

so for the series we have [(-1)^n+2(x-5)^n+1]/[(n+1)5^n+1] all divided by [(-1)^n+1(x-5)^n]/[n5^n]

i simplify and get [-(x-5)n]/[5(n+1)]<1

i am stuck here

 

[d_leet]

well, for the raito test one condition has to be met inorder for the series to converge, that is the abs(an-10/an)<1

so for the series we have [(-1)^n+2(x-5)^n+1]/[(n+1)5^n+1] all divided by [(-1)^n+1(x-5)^n]/[n5^n]

i simplify and get [-(x-5)n]/[5(n+1)]<1

i am stuck here

Well you need to remember that you're taking a limit, and that there is an absolute value so the whole (-1)^k stuff goes away.

 

[Rasine]

ok so i get [(x-5)n]/[5(n+1)]<1 and i need to get a interval of convergence from this

i have no idea

 

[d_leet]

ok so i get [(x-5)n]/[5(n+1)]<1 and i need to get a interval of convergence from this

i have no idea

Did you read everything I said in my last post? You need to take the limit as n goes to infinity.

 

[Rasine]

so [xn-5n]/[5n+5] as n goes to infinity would be [x-5]/5?

 

[d_leet]

so [xn-5n]/[5n+5] as n goes to infinity would be [x-5]/5?

Yes, but in order for the series to converge what are the conditions on that expression.

 

[HallsofIvy]

I don't know why you multiplied that n into x- 5, it's much simpler as
(|x-5|/5)(n/(n+1)). What is the limit of n/(n+1) as n goes to infinity?


(That's simpler to read in LaTex:
\frac{|x-5|}{5}\frac{n}{n+1}
What is the limit of
\frac{n}{n+1}
as n goes to infinity?)

 

[Rasine]

so [x-5]/5<1

and the interval is
-1<[x-5]/5<1
0<x<10

 

[HallsofIvy]

Surely, there's a | key on your keyboard?

 

[Rasine]

thank you for all your help