# What is the convergence radius R of the powerseries sum(n>=1) [z^n / (n*(n+1)) with z complex number?Can we then also show that the powerseries is normal convergent in BR(0) = {z out of C, |z|...

What is the convergence radius R of the powerseries sum(n>=1) [z^n / (n*(n+1)) with z complex number?

Can we then also show that the powerseries is normal convergent in BR(0) = {z out of C, |z| <= R?

### 1 Answer | Add Yours

You need to perform the ratio test such that:

`L = lim_(n-gtoo) [(z^(n+1))/((n+1)(n+2))]*[(n(n+1))/(z^n)]`

Reducing by `z^n` and (n+1) yields:

`L = lim_(n-gtoo) (n*z)/(n+2)`

Since z is not dependent on the limit, you may factor it out of the limit such that:

`L = |z|*lim_(n-gtoo) (n)/(n+2)`

Since the orders of numerator and denominator are alike then the limit is the ratio of leading coefficients such that:

`L = |z|*(1/1) = |z|`

Hence, the ratio test tells us that if `Lgt1` , the series diverges and if `Llt1` , the series converges.

Hence, if `L = |z| lt 1` , the series converges and if `L = |z|gt 1` , the series diverges.

**Hence, evaluating the convergence radius of power series yields R=1.**