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Chapter 12: Problem 11

Find the interval of convergence. $$\sum\left(\frac{k}{100}\right)^{k} x^{k}$$

Short Answer

Expert verified
The interval of convergence is \(\left(-\frac{1}{e}, \frac{1}{e}\right)\).

Step by step solution

01

Apply the Ratio Test

First, we need to apply the Ratio Test, which states that if the limit of \(\frac{a_{n+1}}{a_n}\) as n approaches infinity is L, then: 1. if \(L < 1\), the series converges. 2. if \(L > 1\), the series diverges. 3. if \(L = 1\), the test is inconclusive. In this case, \(a_k = \left(\frac{k}{100}\right)^k x^k\), so we want to find: $$ \lim_{k\to\infty} \frac{a_{k+1}}{a_k} = \lim_{k\to\infty} \frac{\left(\frac{k+1}{100}\right)^{k+1} x^{k+1}}{\left(\frac{k}{100}\right)^k x^k} $$
02

Simplify the Ratio

Now we simplify the ratio expression: $$ \lim_{k\to\infty} \frac{\left(\frac{k+1}{100}\right)^{k+1} x^{k+1}}{\left(\frac{k}{100}\right)^k x^k} = \lim_{k\to\infty} \frac{\left(\frac{k+1}{100}\right)^{k+1}}{\left(\frac{k}{100}\right)^k}x $$
03

Further Simplification

To further simplify the expression, we can rewrite the ratio as: $$ \lim_{k\to\infty} \frac{1}{x} \left(\frac{k+1}{k}\right)^k \left(\frac{k+1}{100}\right) = \lim_{k\to\infty} \frac{1}{x} \left(\frac{k+1}{k}\right)^k \left(\frac{k+1}{100}\right) $$ As we have a known limit in the expression \(\lim_{k\to\infty} \left(\frac{k+1}{k}\right)^k = e\), the ratio becomes: $$ \lim_{k\to\infty} \frac{e}{x} \left(\frac{k+1}{100}\right) $$ To assure convergence, the limit should be less than 1, i.e.: $$\frac{e}{x} \left(\frac{k+1}{100}\right) < 1$$
04

Find the Radius of Convergence

As k goes to infinity, the expression \(\frac{k+1}{100}\) can be neglected, and we are left with: $$\frac{e}{x} \left(\frac{\infty}{100}\right) < 1$$ or $$ \frac{e}{x} <1 $$ We can now find the interval by solving for x. Let the radius of convergence be R: $$R = \frac{1}{e}$$
05

Determine the Interval of Convergence

For the interval of convergence, we consider: $$ -R < x < R $$ Replacing R with its value: $$ -\frac{1}{e} < x < \frac{1}{e} $$ So, the interval of convergence is \(\left(-\frac{1}{e}, \frac{1}{e}\right)\).

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