View Full Version : which series test to use.
rcmango
Jan16-08, 11:42 PM
1. The problem statement, all variables and given/known data
Does this series converge or diverge. n=1 SIGMA infinity ( (n+1)^n / ( n^(n+1) ) )
this could also be changed to lim n-> infinity (1 + 1/n)^n , but then i ask, where the n+1 in the original equation has went?
2. Relevant equations
3. The attempt at a solution
HallsofIvy
Jan17-08, 06:27 AM
Since those are all positive numbers, I would be inclined to use the root test:
[tex]^n\sqrt{\frac{(n+1)^n}{n^{n+1}}}= \frac{n+1}{n^\frac{n+1}{n}}[/itex]
If the limit of that is less than 1, then the series converges.
As to "where did the n+1 go", how did you get "lim (1+ 1/n)^n"?
The summand can be expressed as \frac{ \left( 1 + \frac{1}{n} \right)^n }{n}, but that doesn't really help anyway.
Halls, the root test returns 1, ie inconclusive. I haven't gone through with the calculations but I would try the ratio test next.
The summand can be expressed as \frac{ \left( 1 + \frac{1}{n} \right)^n }{n}, but that doesn't really help anyway.
expressing the summand as \frac{ \left( 1 + \frac{1}{n} \right)^n }{n} does help, you just have to give up finding a test but consider finding a divergent minorante.
\frac{1}{n} < \frac{ \left( 1 + \frac{1}{n} \right)^n }{n} and we know that
\sum_{n=1}^{\infty} \frac{1}{n} = \infty
Damn that is right >.< good work dalle!
rcmango
Jan20-08, 07:32 PM
Okay, so this problem should be approached by the ratio test. We know it diverges, and i believe so because 1/n is a harmonic series.
also, dalle, it looks though that may be similiar to the comparison test then?
and "As to "where did the n+1 go", how did you get "lim (1+ 1/n)^n"?" it was a hint given by the problem and it also is equal to e.
i'm still confused by this.
thankyou for all the help so far.
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