Does the Root Test Determine Convergence for \(\sum \frac{5^n}{n+1}\)?

[srfriggen]

Homework Statement

I don't have the problem in front of me but it was something like "converge or diverge"?:

\sum 5^n/(n+1)

The Attempt at a Solution

I would like to know that if I use the root test, would I get lim n-> \infty 5/(n+1)^1/n = 5/(n+1)^0 = 5/1 = 5 ?

I suspect this is not correct since (n+1)^1^n is an indeterminate form, however, in the book the only way to solve the problem that led to the correct answer was if (n+1) became 1

(this is not the exact example, I am more interested in if you can use the root test on variables variables of n in the base, ex lim -> \inftyn^1/n




If this is not clear enough I will re-post with the correct problem.

 

[Dick]

(n+1)^(1/n) may be indeterminant, but the limit as n->infinity of (n+1)^(1/n) is 1. Take the log and apply l'Hopital.

 

[srfriggen]

do you mean, y=(n+1)^1/n
lny = ln(n+1)^1/n,
lny = (n+1)/n

y'/y = 1/1
e^y'=e^y... ok I got kinda lost

 

[Dick]

do you mean, y=(n+1)^1/n
lny = ln(n+1)^1/n,
lny = (n+1)/n

y'/y = 1/1
e^y'=e^y... ok I got kinda lost

ln((n+1)^(1/n))=(1/n)*ln(n+1)=ln(n+1)/n. Apply l'Hopital. What's limit ln(n+1)/n?

 

[srfriggen]

ln((n+1)^(1/n))=(1/n)*ln(n+1)=ln(n+1)/n. Apply l'Hopital. What's limit ln(n+1)/n?

ok, but when I take the derivative of the top and bottom with respect to n I get (1/(n+1)) / 1 = 1/(n+1), and that -> 0 as n - > infinity. Not sure how you are getting 1 as an answer.

 

[Dick]

ok, but when I take the derivative of the top and bottom with respect to n I get (1/(n+1)) / 1 = 1/(n+1), and that -> 0 as n - > infinity. Not sure how you are getting 1 as an answer.

That limit is telling you that log(y)=0. Remember, you took the log? What's y?

 

[srfriggen]

That limit is telling you that log(y)=0. Remember, you took the log? What's y?

so take e on both sides and you get y = e^0, or y = 1

:)


Thanks! I think that's just something worth memorizing!