Question

Find all values of \(x\) for which the series converges. For these values of \(x,\) write the sum of the series as a function of \(x\). $$ \sum_{n=1}^{\infty}\left(\frac{x^{2}}{x^{2}+4}\right)^{n} $$

Short answer

The series converges for -2 < \(x\) < 2 and the sum of the series, when it converges, is \(\frac{x^{2}}{4}\).

Step-by-step solution

Identify the ratio

In this problem, \(r = \frac{x^{2}}{x^{2}+4}\) . This is the ratio of the geometric series.

Find values of \(x\) for which the series converges

The series will converge if and only if the absolute value of the ratio is less than 1. So, we need to solve the inequality |\(r\)| < 1, i.e., \(\left|\frac{x^{2}}{x^{2}+4}\right| < 1\). This simplifies to -1 < \(\frac{x^{2}}{x^{2}+4}\) < 1. Solving this inequality yields -4 < \(x^{2}\) < 4, which in turn gives -2 < \(x\) < 2. So, the values of \(x\) for which the series converges are all real numbers between -2 and 2 (excluding -2 and 2 itself).

Function for the sum of the series

The sum of a geometric series is given by \(\frac{a}{1 - r}\), where \(a\) is the first term and \(r\) is the ratio. Here, the first term \(a\) is \(\frac{x^{2}}{x^{2}+4}\), and the ratio \(r\) is also \(\frac{x^{2}}{x^{2}+4}\). Thus, the sum \(S\) of the series can be expressed as \(S = \frac{\frac{x^{2}}{x^{2}+4}}{1 - \frac{x^{2}}{x^{2}+4}} = \frac{x^{2}}{4}\). Note that this is valid only when \(x\) is between -2 and 2 (exclusive).