FORM GR1268 Question 18

Question:
Let \(f\) be the function defined by \(\displaystyle f(x)=\sum_{n=1}^\infty\frac{x^n}{n}\) for all \(x\) such that \(-1\lt x\lt 1\). Then \(f'(x)=\)

(A) \(\displaystyle\frac{1}{1-x}\)     (B) \(\displaystyle\frac{x}{1-x}\)     (C) \(\displaystyle\frac{1}{1+x}\)     (D) \(\displaystyle\frac{x}{1+x}\)     (E) \(0\)

Answer:
(A)

Answer Key:
We need the following conditions for term-by-term differentiation:
(1) The series converges at some point \(x_0\); and
(2) The series of derivatives uniformly converges on \((-1,1)\).

The first condition is easy to check.
Take \(x_0=0\).
Then \(f(0)=0\), so it converges to \(0\).

To check the second condition, consider
\[\sum_{n=1}^\infty\left(\frac{x^n}{n}\right)'=\sum_{n=1}^\infty\frac{nx^{n-1}}{n}=\sum_{n=1}^\infty x^{n-1}.\]
This is a geometric series with initial term \(1\) and ratio \(x\), so
\[\sum_{n=1}^\infty x^{n-1}=\frac{1}{1-x}.\]

Note on the sum of geometric series:

Let \(\displaystyle S=\sum_{k=1}^n ar^{k-1} = a + ar + ar^2 + \cdots+ar^{n-1}\).

Then \(\displaystyle rS=\sum_{k=1}^n ar^k = ar +ar^2+ar^3+\cdots+ar^n\).

Subtract the second equation from the first equation:
\[\begin{align}
S-rS &= a-ar^n\\
(1-r)S&=a(1-r^n)\\
S&=\frac{a(1-r^n)}{1-r}.
\end{align}\]
If \(|r|\lt1\), let \(n\rightarrow\infty\) to obtain \(\displaystyle S=\frac{a}{1-r}\).