Power series

Question

Consider the power series
∑n=1∞[(−5)/sqrt(n)]*(x+7)^n.

Find the radius of convergence R.R. If it is infinite, type "infinity" or "inf".

What is the interval of convergence?

Answer 1

Mariam,

 

Let me start by showing the series a little more cleanly:  ∑_n=1^∞ (-5)/√n (x+7)ⁿ.

 

From this you can see that the center of the series is a = -7 (set x+7 = 0 and solve for x).

 

To find the radius of convergence, use the Ratio Test with c_n = (-5)/√n (x+7)ⁿ. a_n+1 = (-5)/√(n+1) (x+7)ⁿ⁺¹.

 

lim_n→∞ |a_n+1/a_n| = lim_n→∞ √(n+1)/√n |x+7| (after simplifying)

                              = |x+7| lim_n→∞ √[(n+1)/n]

                              = |x+7| lim_n→∞ √(1 + 1/n)

                              = |x+7| √(1 + 0) = |x+7|.

 

In order for the series to converge (absolutely), |x+7| < 1. You must solve this for x and test convergence at the endpoints of the interval of solution.

 

-1 < x+7 < 1

-8 < x < -6

 

Since the center is a = -7 and the interval extends one unit to the left and right of -7, the radius of convergence is R = 1.

 

If you test convergence at the endpoints (separately), by plugging in the values and analyzing the series, you will see the following:

 

x = -8  ⇒  ∑_n=1^∞ (-5)/√n (-1) = 5 ∑_n=1^∞ 1/√n

This is 5 times the p-series with p = 1/2 (√n = n^1/2), so must be divergent.

 

x = -7  ⇒  ∑_n=1^∞ (-5)/√n (1) = -5 ∑_n=1^∞ 1/√n

This is -5 times the same divergent p-series, so is also divergent.

 

To sum up, the series converges absolutely in the interval (-8,-6), and diverges at x = -8 and x = -7, so the interval of convergence is (-8,-7).