Expert: Paul Klarreich - 2/5/2008

Question
does the series from 1 to infinity of (1 - (2/n))^3n convegre or diverge.

i've tried almost all the tests and cant figure it out....

Answer
Questioner:   sean
Category:  Calculus
Private:  No
 
Subject:  series converge or diverge
Question:  does the series from 1 to infinity of (1 - (2/n))^3n convegre or diverge.

I've tried almost all the tests and can't figure it out....
.......................................
Hi, Sean,

This is your series.

inf
SUM (1 - 2/n)^3n
n=1  

I think this series DIVERGES.  When n > 2, the terms will all be positive.  A requirement for a series of positive terms to converge is that  lim a[n] = 0.  I.e. if the terms themselves don't approach zero, the sum can't converge.
So what about the terms:

lim (1 - 2/n)^3n

Let k = n/2, then  n = 2k, and this is:

lim (1 - 1/k)^6k =

lim [(1 - 1/k)^k]^6 =

[lim (1 - 1/k)^k ]^6

But look at the expression inside the [].  That limit is 1/e.
[Proof upon request.]
[e = 2.7178..., which I think you knew.]

So the terms approach  1/e^6, which ain't zero.  [Small, but not zero.]