Question

How do you find the radius of convergence `Sigma (n^nx^n)/(lnn)^n` from `n=[2,oo)`?

Answer 1

The root test states that the series `suma_n` is convergent if `L=lim_(nrarroo)root(n)abs(a_n)<1`.

Here we see that `a_n=((nx)/lnn)^n`, so:

`L=lim_(nrarroo)root(n)abs(((nx)/lnn)^n)=lim_(nrarroo)abs((xn)/lnn)`

Since the limit is dependent only on the change in `n`, the `x` can be removed from the limit like a constant:

`L=absxlim_(nrarroo)abs(n/lnn)`

We see that the limit approaches `oo`, since `n` grows faster than the logarithmic function `lnn`. The only time when `L<1`, when the series converges, will be when `L=0`, which occurs only when `x=0`.

Since there is only one value of `x` for which the series converges, we see that `R=0`.