Question

The power series \(\sum_{n=0}^{\infty} a_{n} x^{n}\) converges for \(|x+1|<4\) What can you conclude about the series \(\sum_{n=0}^{\infty} a_{n} \frac{x^{n+1}}{n+1} ?\) Explain.

Short answer

The series \(\sum_{n=0}^{\infty} a_{n} \frac{x^{n+1}}{n+1}\) converges on the interval \(-5 \leq x \leq 3\), the same as the original series.

Step-by-step solution

Identify Inputs and Modifications

In the new series, \(x^{n}\) is replaced with \(\frac{x^{n+1}}{n+1}\). This means x is multiplied by an additional term, and additional term is dividing the series. Whilst it's easy to see that nothing has changed about the overall 'shape' of the values that x can take (\(-5 \leq x \leq 3\), it may be necessary to consider how the additional division of the \(a_{n}\) terms by (n+1) could impact convergence.

Analyze New Series

Since the coefficients of the series, \(a_{n}\), are only divided by (n+1), and the exponents of x are increased by 1, the overall effect is that the terms of the series are made somewhat smaller (non-increasing) compared to the original series. It can be concluded that for any x within the radius of convergence of the original series, the new series also converges.

State the Conclusion

Given the overall effect of the modification of the original series to our new series, we can conclude that the new series, \(\sum_{n=0}^{\infty} a_{n} \frac{x^{n+1}}{n+1} \), must also converge for values of x in the interval \(-5 \leq x \leq 3\).