Convergence of Series: Ratio Test vs. Comparison Test

[rcmango]

Homework Statement

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?

Homework Equations

The Attempt at a Solution

 

[HallsofIvy]

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]<br /> If the limit of that is less than 1, then the series converges.<br /> <br /> As to "where did the n+1 go", how did you get "lim (1+ 1/n)^n"?[/tex]

 

[Gib Z]

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.

 

[dalle]

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
$$\sum_{n=1}^{\infty} \frac{1}{n} = \infty$$

 

[Gib Z]

Damn that is right >.< good work dalle!

 

[rcmango]

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 similar 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.