- #1

tolove

- 164

- 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}= -1/3raise both sides

**a**

_{n}= e^{(-1/3)}Which would mean divergence, right? Since the nth term does not equal 0?

But this problem converges