# Convergence problem (nth-term test)

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.

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

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

But this problem converges

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LCKurtz
Homework Helper
Gold Member
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.

an = (n/(3n+1))n
Take ln of both sides, then use n = 1/(1/n) to setup for l'Hopital's rule.
ln an = ln(n/(3n+1)) / (1/n)
l'Hop
But that is neither an ##\frac\infty \infty## or ##\frac 0 0## form.

But that is neither an ##\frac\infty \infty## or ##\frac 0 0## form.
Ohh, ok

1) If $\sum$ an converges, then an → 0

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

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

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

LCKurtz
Homework Helper
Gold Member
Ohh, ok

1) If $\sum$ an converges, then an → 0

2) $\sum$ an may or may not converge if an → 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.

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.