Convergence problem (nth-term test)

[tolove]

Show that the sum of (n/(3n+1))ⁿ from n=1 to ∞ converges.

The book solves this with a comparison test to (1/3)ⁿ, but I'm making a mistake with an n-th term test somewhere.

a_n = (n/(3n+1))ⁿ
Take ln of both sides, then use n = 1/(1/n) to setup for l'Hopital's rule.
ln a_n = ln(n/(3n+1)) / (1/n)
l'Hop
ln a_n = -n/(3n+1)
l'Hop
ln a_n = -1/3
raise both sides
a_n = e^(-1/3)

Which would mean divergence, right? Since the nth term does not equal 0?

But this problem converges

 

[LCKurtz]

Show that the sum of (n/(3n+1))ⁿ from n=1 to ∞ converges.

The book solves this with a comparison test to (1/3)ⁿ, but I'm making a mistake with an n-th term test somewhere.

a_n = (n/(3n+1))ⁿ
Take ln of both sides, then use n = 1/(1/n) to setup for l'Hopital's rule.
ln a_n = ln(n/(3n+1)) / (1/n)
l'Hop

But that is neither an $\frac\infty \infty$ or $\frac 0 0$ form.

 

[tolove]

But that is neither an $\frac\infty \infty$ or $\frac 0 0$ form.

Ohh, ok

1) If \sum a_n converges, then a_n → 0

2) \sum a_n may or may not converge if a_n → c, c being any real number.

3) If a_n as n → ∞ fails to exist, then \sum a_n diverges.

e: thank you very much, this was driving me nuts

 

[LCKurtz]

Ohh, ok

1) If \sum a_n converges, then a_n → 0

2) \sum a_n may or may not converge if a_n → c, c being any real number.

If $a_n$ does anything but converge to 0, the series diverges. If it does converge to zero, the series may or may not converge. That's all there is to it.

 

[tolove]

If $a_n$ does anything but converge to 0, the series diverges. If it does converge to zero, the series may or may not converge. That's all there is to it.

Double thanks, my mistake is with l'Hopital's rule.

f(a) = g(a) = 0 as the limit of n → a must be true.
The top function:
ln(n/(3n+1))
does not go to 0 as n → infinity, so l'Hopital's rule cannot be applied.