Convergence radius for $\sum_{n=1}^\infty n(-x)^{n-1}$

Question

I have the following multiple choice question.

What is the convergence radius, $\rho$ for $$\sum_{n=1}^\infty n(-x)^{n-1} $$

1. $\rho=0$

2. $\rho = 1$

3. $\rho = -1$

4. $\rho = 2$

5. The sequence doesn't have a convergence radius at all.

My attempt

Call $a_n=n(-x)^{n-1}$. We use the "ratio test". $$\Big| \frac{a_{n+1}}{a_n} \Big| = \Big| \frac{(n+1) \cdot (-x)^n}{n(-x)^{n-1}} \Big|=\Big| \frac{x(n+1)}{n} \Big|$$

We see that $\Big| \frac{x(n+1)}{n} \Big| \rightarrow |x| \: \: \text{for} \: \: n\rightarrow \infty$.

For $x$ to be convergent, we must have $|x|<1$, which finally means that $$-1<x<1 $$ which I assume means that the sequence has a convergence radius of $1$.

However, using Maple to double check, I find that the sequence actually converges when $x>-1$, and there is no upper bound on it. Am I doing something wrong?

Checking with Maple

Answer 1

You asked about Maple. Using the latest version (Maple 2020) one can obtain the following piecewise result, where x is taken as a parameter.

restart;

sum(n*(-x)^(n-1),n=1..infinity,parametric);

           /                     1     
           |      0            - - = 0 
           |                     x     
           |                           
           |      1                    
           |  ---------        |x| < 1 
           |          2                
          <   (-x - 1)                 
           |                           
           |   infinity                
           | - ---------       1 <= -x 
           |   signum(x)               
           |                           
           |  undefined       otherwise
           \                           

As for the results you found strange, for x>=1, notice that the Description in the Maple Help page for numeric summation mentions the concept of a "formal" summation.

If you force that off by supplying the formal=false option then result would be, for example,

restart;

evalf(Sum(n*(-1)^(n-1),n=1..infinity,formal=false));

           Float(undefined)

evalf(Sum(n*(-2)^(n-1),n=1..infinity,formal=false));

           Float(undefined)

You unfortunately wrapped lowercase sum in a call to evalf. The inner call to sum will return with the formal option stripped off, hence the undesired result.

restart;

sum(n*(-1)^(n-1),n=1..infinity,formal=false);

                infinity             
                 -----               
                  \                  
                   )          (n - 1)
                  /     n (-1)       
                 -----               
                 n = 1               

lprint(%);

       sum(n*(-1)^(n-1),n = 1 .. infinity)

evalf(%);

               0.2500000000

Answer 2

You are correct that radius of convergence is 1 (answer choice 2). However the ratio test is inconclusive if the series will converge when $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = 1$. To determine this you need you need to check the end points -1 and 1. In this case, the series diverges at both end points $x=-1$ and $x=1$ so the interval of convergence is $(-1, 1)$.