Converge or diverge?
$$\sum_{n=1}^\infty \dfrac{log(n+1)-log n}{tan^{-1}(2/n)}$$
Please, help

Question

amistre64 · Answer

im wondering if a ratio test give us anything conclusive?

it appears the top goes to 0, and the bottom goes to 0  might be lhopable?

loser66 · Answer

Apply l'hospital: the summand is
numerator: (1/(n+1) - (1/n) = -1/(n^2+n)
denominator : (1/1+4n^2)
combine:
$$-\dfrac{1+4n^2}{n^2+n}$$
hence the lim =-4 $\neq 0$ divergence, right?

amistre64 · Answer

ill have to look that up to be sure; but memory wants to say: if the limit doesnt go to zero for the terms, it diverges. http://www.mathwords.com/l/limit_test_for_divergence.htm

amistre64 · Answer

the wolf says the terms limit to 1/2 http://www.wolframalpha.com/input/?i=limit%28n+to+inf%29+%28ln%28n%2B1%29-ln%28n%29%29%2F%28tan%5E%28-1%29%282%2Fn%29%29

amistre64 · Answer

and by the limit test, it diverges .. as we expected.

loser66 · Answer

Oh yea, I messed up. thank you.