How do you find the radius of convergence of #sum_(n=0)^oox^n# ?

Question

Isaiah Allen · Accepted Answer

By Ratio Test, we can find the radius of convergence: #R=1#. By Ratio Test, in order for #sum_{n=0}^{infty}a_n# to converge, we need
#\lim_[n to infty}|{a_{n+1}}/{a_n}|<1#. For the posted power series, #a_n=x^n# and #a_{n+1}=x^{n+1}#. 
So, we have
#\lim_[n to infty}|{x^{n+1}}/{x^n}|=lim_{n to infty}|x|=|x|<1=R# Hence, its radius of convergence is #R=1#.

Evelyn Agard · Answer

undefined To find the radius of convergence of the series $ \sum_{n=0}^\infty x^n $, where $ x $ is a real or complex number, you can use the ratio test:

1. Apply the ratio test:
$$ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{x^{n+1}}{x^n} \right| = \lim_{n \to \infty} |x| $$

2. Determine the convergence behavior based on the value of $ |x| $:
   - If $ |x| < 1 $, the series converges absolutely.
   - If $ |x| > 1 $, the series diverges.
   - If $ |x| = 1 $, the test is inconclusive, and we need to investigate further.

3. Thus, the radius of convergence, denoted by $ R $, is the distance from the center of the series (which is 0 in this case) to the nearest point at which the series converges absolutely. For the series $ \sum_{n=0}^\infty x^n $, $ R = 1 $.