Radius of Convergence

Definition

• Given ${\sum }_{n=1}^{\infty }{C}_n(x=a{)}^n$
• The largest $R\in {\mathbb{R}}^{\ge 0}\cup \{\infty \}$ s.t increasing power series
• We Absolute Convergence when $|x-a|<R$
• We Diverge when $|x-a|>R$ This $R$ is called the radius of convergence for the power series

Finding Radius of Convergence Process

From a given Series ${\sum }_{n=1}^{\infty }{a}_n$

1. Simplify the radius formula which is $|\frac{a_{n+1}}{a_n}|$
2. Find the interval $I$ where $|\frac{a_{n+1}}{a_n}|<1$
3. For those endpoints on the interval $I$, check for the values of $x$ in the original ${\sum }_{n=1}^{\infty }{a}_n$ formula with another test to check for divergence or convergence
4. The final interval is the interval of convergence.

Example

Find the radius of convergence of ${\sum }_{n=1}^{\infty }\frac{a(x+2{)}^n}{3^{n+1}}$

Soln

Step 1

Find $I$

• With $a=-2$, $C_n=\frac{n}{3^{n+1}}$ Find $R$
• ${\lim }_{n\to \infty }|\frac{C_{n+1}(x-a{)}^{n+1}}{C_n(x-a{)}^n}|$
• $={\lim }_{n\to \infty }|\frac{n-1}{3^{n+2}}(x+2{)}^{n+1}||\frac{n(x+2{)}^n}{3^{n+1}}|$
• $={\lim }_{n\to \infty }\frac{n+1}{n}\ast \frac{3^{n+1}}{3^{n+2}}\ast |\frac{(x+2{)}^{n+1}}{(x+2{)}^n}|$
• $={\lim }_{n\to \infty }(1+\frac{1}{n})(\frac{1}{3})(x+2)$
• $=\frac{1}{3}|x+2|$ By Ratio Test, our Power Series:
• Will AC when $\frac{1}{3}|x+2|<1$
• $⟹|x+2|<3$
• Will Diverge when $\frac{1}{3}|x+2|>1$
• $⟹|x+2|>3$ By definition of $R$, we get $R=3$

Step 2

For step 2, we check the endpoints for $x=a\pm R$ . Plug them into the power series, and get a numerical series

• Check $x=a\pm R=-5,1$
• For $x=-5$:
  • We get ${\sum }_{n=1}^{\infty }\frac{(n)(-5+2{)}^n}{3^{n+1}}$
  • $={\sum }_{n=1}^{\infty }\frac{(n)(-3{)}^n}{3\ast 3^n}$
  • $={\sum }_{n=1}^{\infty }\frac{(n)(-1{)}^n(3{)}^n}{3\ast 3^n}$
  • $={\sum }_{n=1}^{\infty }\frac{(n)(-1{)}^n}{3}$
  • Applying div-test for ${\lim }_{n\to \infty }\frac{(-1{)}^n(n)}{3}=\pm \infty \ne 0$
  • Thus, by Div Test, our power series diverges at $x=-5$
• For $x=1$
  • We get ${\sum }_{n=1}^{\infty }\frac{(n)(1+2{)}^n}{3^{n+1}}$
  • $={\sum }_{n=1}^{\infty }\frac{(n)(3{)}^n}{3\ast 3^n}$
  • $={\sum }_{n=1}^{\infty }\frac{(n)}{3}$
  • Applying div-test for ${\lim }_{n\to \infty }\frac{n}{3}$, our power series diverges. Thus, our final interval $I=(-5,1)$
• We have Absolute Convergence (and by proxy, Convergence between those values
• We have Divergence outside that interval