Let \(f\) be analytic in \(|z|<1\) and suppose \[ |f(z)|\le\frac1{1-|z|}. \] Show that, for \[ a_n=\frac{f^{(n)}(0)}{n!}, \] one has \[ |a_n|\le (n+1)\left(1+\frac1n\right)^n<e(n+1). \]
Solution
The displayed estimate is for integers \(n\geq1\), as indicated by the occurrence of \(1/n\).
1. For every \(0<r<1\) and every integer \(n\geq0\),\[ \abs{a_n} \leq \frac1{(1-r)r^n}. \]
By Cauchy’s coefficient formula on the circle \(\abs z=r\), \[ a_n = \frac1{2\pi i} \int_{\abs z=r} \frac{f(z)}{z^{n+1}}\,dz. \] The hypothesis gives \[ \max_{\abs z=r}\abs{f(z)} \leq \frac1{1-r}. \] Therefore \[ \begin{aligned} \abs{a_n} &\leq \frac1{2\pi} \cdot 2\pi r \cdot \frac1{1-r} \cdot \frac1{r^{n+1}}\\ &= \frac1{(1-r)r^n}. \end{aligned} \]
2. For every integer \(n\geq1\),\[ \abs{a_n} \leq (n+1)\left(1+\frac1n\right)^n. \]
In step 1 choose \[ r=\frac{n}{n+1}. \] Then \[ 1-r=\frac1{n+1} \] and \[ r^{-n} = \left(\frac{n+1}{n}\right)^n = \left(1+\frac1n\right)^n. \] Substitution gives the asserted inequality.
3. For every integer \(n\geq1\),\[ \left(1+\frac1n\right)^n<e. \]
For every \(x>0\), \[ \log(1+x)<x. \] Indeed, the function \[ h(x)=x-\log(1+x) \] satisfies \[ h(0)=0, \qquad h'(x)=\frac{x}{1+x}>0 \] for \(x>0\).
Taking \(x=1/n\) gives \[ n\log\left(1+\frac1n\right)<1. \] Exponentiating yields the claim.
4. Hence, for every integer \(n\geq1\),\[ \boxed{ \abs{a_n} \leq (n+1)\left(1+\frac1n\right)^n < e(n+1) }. \]
Combine steps 2 and 3.
5. Q.E.D.
Step 4 is the required estimate.