Suppose I have a series, say SUM(3/((2n + 1)[sup]1/n[/sup])) from 1 to infinity. Now we know that this series converges by p series; how can I find the value that the limit converges to? This is all stuff I should know, but it's been so long.


edit: suppose the series is SUM(1/e[sup]n[/sup]) 1 to infinity. I like this better than the previous example. This is a gemetric series, so my first thought was 1/(1 - e); this gives the correct answer only negative. Hooh?

You're wrong about the first series; it diverges to infinity.

About the second:
Sum [n=1 to infinity] ((1/e)^n)
Is an infinite converging geometric series.
The first element is: A1 = 1/e
The quotient is: q = 1/e
Therefore the sum is:
S = A1 / (1 - q) = (1/e) / (1 - 1/e) = 1 / e(1 - 1/e) = 1 / (e - 1)

[quote author=Yoni link=board=36;threadid=5550;start=0#msg47144 date=1078273626]
You're wrong about the first series; it diverges to infinity.
[/quote]

Oops - you're right. I got confused. The sequence {3/((2n + 1)[sup]1/n[/sup]) } converges. (To 3, I think)

Edit: that means the series diverges by the Nth term test.

Correct. Short proof:

Two basic limits are [color=yellow]n^(1/n) -> 1[/color] and [color=yellow]c^(1/n) -> 1[/color] for constant c > 0.
From this, by multiplying, you get, [color=yellow](2n)^(1/n) -> 1*1 = 1[/color] and [color=yellow](3n)^(1/n) -> 1*1 = 1[/color].
Using the Sandwich rule, you get [color=yellow](2n + 1)^(1/n) -> 1[/color], therefore [color=yellow]3/((2n + 1)^(1/n)) -> 3/1 = 3[/color].

This is also a proof that the matching series doesn't converge. For a series Sum(A(n)) to converge, it must satisfy A(n) -> 0. (Edit: You just said this in your edit. :))