Problem 10 - Calculus Early Transcendentals

Question

Determine the radius of convergence of the following power series. Then test the endpoints to determine the interval of convergence. $$\sum \frac{(2 x)^{k}}{k !}$$

Step-by-Step Solution

Answer

Answer: The interval of convergence for the given power series is \((-\infty, \infty)\).

Step 1: Apply the ratio test

For a power series of the form $$\sum a_k (x - c)^k$$, we will use the ratio test to determine the radius of convergence: $$R = \lim_{k\to\infty} \left|\frac{a_k}{a_{k+1}}\right|$$ For our given power series $$\sum \frac{(2x)^k}{k!}$$, we have \(a_k = \frac{(2x)^k}{k!}\), so let's find the limit: $$\lim_{k\to\infty} \left|\frac{\frac{(2x)^k}{k!}}{\frac{(2x)^{k+1}}{(k+1)!}}\right|$$

Step 2: Simplify the limit expression

We can simplify the expression inside the limit by multiplying by the reciprocal of the denominator: $$\lim_{k\to\infty} \left|\frac{\frac{(2x)^k}{k!} \cdot \frac{(k+1)!}{(2x)^{k+1}}}{1}\right|$$ Now we can cancel out common terms: $$\lim_{k\to\infty} \left|\frac{(2x)^k \cdot (k+1)!}{k!\cdot (2x)^{k+1}}\right| = \lim_{k\to\infty} \left|\frac{(2x)^k}{(2x)^{k+1}} \cdot \frac{(k+1)!}{k!}\right|$$ This simplifies to: $$\lim_{k\to\infty} \left|\frac{1}{2x} \cdot (k+1)\right|$$

Step 3: Evaluate the limit and find the radius of convergence

Since the limit is infinity when x is nonzero, the series converges for all x values: $$R = \frac{1}{2x} \cdot (k+1)$$

Step 4: Test the endpoints of the interval of convergence

The power series converges when \(R > 0\), so let's test the endpoints of the interval \((-\infty, \infty)\) by plugging in \(x = -\infty\) and \(x = \infty\): For \(x = -\infty\), \(R = \frac{1}{-2\infty} \cdot (k+1) = 0\). For \(x = \infty\), \(R = \frac{1}{2\infty} \cdot (k+1) = 0\). Since the series converges for all values of x, the interval of convergence is \((-\infty, \infty)\).