Derivative Of A Power Series Has The Same Radius Of Convergence

Consider the two series:

\[\sum^\infty_{n=0} a_n(x-a)^n \qquad \sum^\infty_{n=1} (n)a_n(x-a)^{n-1} \]

Let \(x-a=z\):

\[\sum^\infty_{n=0} a_n z^n \qquad \sum^\infty_{n=1} (n)a_n z^{n-1} \]

Assume that \(\sum^\infty_{n=0} a_n z^n\) has a radius of convergence \(R\). For any \(z\) you choose, you can also think of a positive integer \(p\) such that \(|z| < p < R\). Since \(|z| < p\):

\[0 < \frac{|z|}{p} < 1\]

We can rewrite \(|n a_n z^{n-1}|\) as:

\[\begin{align} |n a_n z^{n-1}| &= \left| \frac{n a_n z^{n-1}p^n}{p^n} \right| = \left| \frac{n a_n (z)^{n-1}p^n}{p^{n-1}p} \right| \\ &= \left| \frac{n a_n p^n}{p} \right| \left( \frac{|z|}{p} \right)^{n-1} = \frac{n}{p} \left( \frac{|z|}{p} \right)^{n-1} |a_n p^n| \end{align}\]

Let \(r = \frac{|z|}{p}\):

\[|n a_n z^{n-1}| = \frac{n r^{n-1}}{p} |a_n p^n| \]

The ratio test shows that the series \(\sum n r^{n-1}\) converges:

\[\lim_{n \to \infty} \frac{(n+1) r^{n}}{n r^{n-1}} = \lim_{n \to \infty} \left( 1+ \frac{1}{n} \right) r < 1 \]

Let \(M\) be an upper bound of sequence \(\{ n r^{n-1} \}\):

\[\begin{align} \frac{n r^{n-1}}{p} |a_n p^n| &\le \frac{M}{p} |a_n p^n| \\ |n a_n z^{n-1}| &\le \frac{M}{p} |a_n p^n| \end{align}\]

Earlier we assumed that \(\sum^\infty_{n=0} a_n z^n\) has a radius of convergence \(R\). If \(\sum^\infty_{n=0} a_n z^n\) converges when \(z = b\), then \(\sum^\infty_{n=0} a_n z^n\) converges absolutely for all \(|z| \lt b\), this was explored here. This means \(\sum^\infty_{n=0} | a_n z^n |\) converges for all \(-R < z < R\).

Since \(p < R\), then \(\sum^\infty_{n=0} |a_n p^n|\) converges. By the inequality above and the comparison test, we can claim \(\sum^\infty_{n=0} |n a_n z^{n-1}|\) is also convergent, and thus so is \(\sum^\infty_{n=0} n a_n z^{n-1}\).

If \(|z| > R\), then \(\sum^\infty_{n=0} a_n z^n\) diverges. Therefore, \(\left[ \frac{1}{z}\sum^\infty_{n=0} a_n z^n \right]\) diverges.

\[\begin{align} (n) a_n z^{n-1} &\ge a_n z^{n-1} \\ n a_n z^{n-1} &\ge \frac{1}{z} a_n z^n \end{align}\]

By the comparison test, if \(\left[ \sum^\infty_{n=0} \frac{1}{z} a_n z^n \right]\) diverges then \(\sum^\infty_{n=1} n a_n z^{n-1} \) diverges as well.

Styles

(uses cookies)