MHB Please check this convergence test (#2)

[ognik]

$ \sum_{n}\frac{1}{n.{n}^{\frac{1}{n}}} $

Now $\frac{1}{n}$ diverges and $\ne 0$ , so by limit comparison test:

$ \lim_{{n}\to{\infty}} \frac{n.{n}^{\frac{1}{n}}}{n} = \lim_{{n}\to{\infty}} {n}^{\frac{1}{n}} = \lim_{{n}\to{\infty}} {n}^0 = 1$ (I think the 2nd last step may be dubious?)

Therefore both series diverge

(also let me know if there is another approach, thanks)

 

[GJA]

Hey (again) ognik,

It is true that $\lim_{n\rightarrow\infty}n^{\frac{1}{n}}=1$, but not for the reason you've stated (i.e. $n^0=1$). The limit $\lim_{n}n^{\frac{1}{n}}$ is an "indeterminate form" from calculus. The trick is to write $y=n^{\frac{1}{n}}$, take a logarithm on both sides, then evaluate the limit using L'Hopital's rule to get $\lim_{n}\ln(y)=0\Longrightarrow\lim_{n}y=1.$

 

[ognik]

Thanks, slowly re-learning all these tricks of the trade...

$ \lim_{{ n}\to{\infty }}ln (y) = \lim \frac{ln (n)}{n} = \lim \frac{1}{n} = 0; \therefore \lim y = e^{0} = 1$