Question

The radius of convergence of the power series \(\sum_{n=0}^{\infty} a_{n} x^{n}\) is \(3 .\) What is the radius of convergence of the series \(\sum_{n=1}^{\infty} n a_{n} x^{n-1} ?\) Explain.

Short answer

The radius of convergence of the series \(\sum_{n=1}^{\infty} n a_{n} x^{n-1}\) is 3.

Step-by-step solution

Understanding the given information

Initially the problem provides a power series \(\sum_{n=0}^{\infty} a_{n}x^{n}\) and states that the radius of convergence is 3. It then asks about the radius of convergence of the series \(\sum_{n=1}^{\infty} n a_{n} x^{n-1}\). This new series is derived from the original series by multiplying each term by \(n\) and decreasing the power of \(x\) by 1.

Observing the effects of the manipulations

Neither multiplying each term by \(n\) nor decreasing the power of \(x\) by 1 affects the radius of convergence. The radius of convergence is generally determined by the coefficients \(a_n\) of the series and is not affected by the indexing or the power of \(x\). Therefore, even though the second series is different due to the multiplication by \(n\) and the change of power of \(x\), the radius of convergence remains the same as in the first series.

Result

The radius of convergence of the series \(\sum_{n=1}^{\infty} n a_{n} x^{n-1}\) is the same as the radius of convergence of the original series, which is 3.