Yes, if the limit of the sequence of terms is not 0 the series does not converge. In order for the series to converge, the sequence of terms must converge to 0fast enough. For example, the terms converge to 0 but the series does NOT converge, by the "integral test". In a sense, that is the "critical example"- the series converges if p> 1, diverges if .

That's not bad but simpler is to use the "ratio test". If then so that which goes to 0 as n goes to infinity. A (non-negative) series will converge if and only if the sequence [itex]a_{n+1}/a_n[/tex] converges to a number less than 1.I've been solving the problems very awkwardly and usually incorrectly.

For example:

Infinite series:

n! / 6^n

My logic is that it must diverge simply because for very large n, n! becomes greater than 6^n, even though that's not the case at lower n.

[quote]For the infinite sum (2 + (3/n))^n, I just logically concluded that 2 does not matter at large n, than 3/n approaches 0, and (3/n)^n just approaches 0 faster, so in my mind the series has to converge. Something tells me that is not the case, though.[quote]

That 2 certainlydoesmatter! Yes, 3/n approaches 0 which mean the sum basically involves adding 2 to itself an infinite number of times. Since you had already said that you understand that if the sequence of terms of a sum does not go to 0, the sum itself does not converge. Ok, does NOT go to 0 as n goes to infinity.

Yes, a sum can still diverge if the terms approach 0. As I said before, they have to go to 0 "fast enough"- faster then 1/n. However, your thinking that goes to 0 is completely wrong! For large n, 3/n goes to 0 so this is close to [tex]2^n[/itex] which goes to infinity, not 0!I believe my that my thinking is fine for the first, but wrong for the second. Can it still diverge if terms approach 0 as n grows without bound? An example and why, would be great!

Thank you!