diverge or converge?

Question

E from 2 to infinity
3/ ((n^2)-2)
this series converge or diverge
a) using direct comparision
b)using limit comparision

Answer 1

I assume that by "E" you mean "Sigma" (Σ). Notice that every term in this series is positive and each is less than 3/(n^2).

a) The direct comparison test says that if you treat these two series as functions, the integral of the first (from 2 to infinity) converges if the integral of 3/(n^2) converges. The integral of this is (-3n^(-1)), which when you plug in the bounds gives you (0 - (-3(2)^-1)) = 3/2. So the original series converges too.

b) The limit comparison test says that if you have two continuous functions whose ratio has a limit, then they either both converge or both diverge.
lim n→∞ [3/((n^2)] / [3/((n^2) - 2)] =
lim n→∞ [(n^2) - 2] / n^2 =
lim n→∞ 1 - (2/ n^2) = 1
We know 3/n^2 converges, so the other limit does too.

Answer 2

Direct comparisson doesn't work here, because for every n>=2, 1/(n^2 -2) > 1/n^2. So, though Sum 1/n^2 converges, we don't conclude anything.

But the limit comparisson test works: lim (1/(n^2 -2))/(1/n^2) = lim n^2/(n^2-2) = 1 >0. So, the 2 series converge or diverge. But since Sum (1/n^2) converges, then Sum (1/(n^2 -2)), and therefore, Sum (3/(n^2 -2)) converge.

Answer 3

It's going to converge absolutely at zero because n^4 is less than the absolute value of (-2)^n for n > 16. So, as n approaches infinity, the result of your equation will near zero.

Answer 4

It looks like n^-2, so converges. the 3 and the 2 ill not matter.

Answer 5

3/((n^2)-2) =< c/n^2, with c>0.
and sum(c/n^2, n=2, inf) converge (=c*pi^2/6)
=> series converge

Answer 6

Ans
Clearly this series converges as n tends to infinity