Question

Find the values of \(x\) for which the infinite series \(1+2 x+x^{2}+2 x^{3}+x^{4}+2 x^{5}+x^{6}+\cdots\) converges. What is the sum when the series converges?

Short answer

The series \(1+2x+x^{2}+2x^{3}+x^{4}+2x^{5}+x^{6}+ \dots\) converges for \(x\) in the interval \(-1 < x < 1\). When it does converge, the sum is \((1 + 2x)/(1 - x^{2})\).

Step-by-step solution

Identify the pattern of the sequence

By examining the series \(1+2x+x^{2}+2x^{3}+x^{4}+2x^{5}+x^{6}+ \dots\), we can identify a pattern. We see that the series is alternating between coefficients of 1 and 2, and the exponent on \(x\) is increasing by 1 for each term. This can be separated out into two different series: one with the coefficients of 1, which would be \(1 + x^{2} + x^{4} + x^{6} + \dots\) and one with the coefficients of 2, which would be \(2x + 2x^{3} + 2x^{5} + \dots\).

Find the common ratio in each series

For the first series, the common ratio is \(x^{2}\). For the second series, we can factor out the 2 to get \(2(x + x^{2} + x^{4} + \dots)\), which also has a common ratio of \(x^{2}\). Therefore, in both series the common ratio is \(x^{2}\).

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

For a geometric series to converge, the absolute value of the common ratio must be less than 1, i.e., \(|-x^{2}| < 1\). Therefore, \(-1 < x < 1\). So, the series converges for \(x\) within this interval.

Find the sum when the series converges

The sum of an infinite geometric series is given by \(S = a/(1 - r)\), where \(a\) is the first term and \(r\) is the common ratio. For the first series, \(S_{1} = 1/(1 - x^{2})\). For the second series, \(S_{2} = 2x/(1 - x^{2})\). The total sum of the series \(S_{total}\) will be the sum of \(S_{1}\) and \(S_{2}\), i.e., \(S_{total} = 1/(1 - x^{2}) + 2x/(1 - x^{2}) = (1 + 2x)/(1 - x^{2})\). So the sum of the series, when it converges, is \((1 + 2x)/(1 - x^{2})\).