Let R > 1 and let f be analytic on IzI < R except at z = 1, where f has a simple pole. If M n=O is the Maclaurin series for f, show that hn,+,M alL exists.

To show that the limit lim⁡n→∞nan\lim_{n \to \infty} n a_nlimn→∞​nan​ exists for the Maclaurin series f(z)=∑n=0∞anznf(z) = \sum_{n=0}^{\infty} a_n z^nf(z)=∑n=0∞​an​zn of the function $f$, which is analytic in $|z|<R$ except at $z=1$ where $f$ has a simple pole, we can use the residue theorem and the properties of the Laurent series.

Step-by-Step Proof:

1. Analyticity and Singularities:
   • The function $f(z)$ is analytic on $|z|<R$ except at $z=1$, where it has a simple pole.
   • Near $z=1$, the function $f(z)$ can be expressed as: f(z)=Az−1+g(z),f(z) = \frac{A}{z – 1} + g(z),f(z)=z−1A​+g(z), where $A$ is the residue of $f$ at $z=1$ and $g(z)$ is analytic at $z=1$.
2. Residue Calculation:
   • The residue at $z=1$ is given by A=lim⁡z→1(z−1)f(z)A = \lim_{z \to 1} (z – 1)f(z)A=limz→1​(z−1)f(z).
3. Maclaurin Series Representation:
   • The Maclaurin series of $f(z)$ is: f(z)=∑n=0∞anzn,f(z) = \sum_{n=0}^{\infty} a_n z^n,f(z)=n=0∑∞​an​zn, valid for $|z|<R$.
4. Behavior of ana_nan​:
   • Because $f(z)$ has a simple pole at $z=1$, the Laurent series expansion around $z=1$ includes a term Az−1\frac{A}{z – 1}z−1A​.
   • For $|z|<1$, the Maclaurin series coefficients ana_nan​ are related to the behavior of $f(z)$ near $z=0$.
5. Analyzing the Laurent Series and Maclaurin Series:
   • The Laurent series expansion around $z=1$ can be written as: f(z)=Az−1+g(z),f(z) = \frac{A}{z – 1} + g(z),f(z)=z−1A​+g(z), where $g(z)$ is analytic at $z=1$.
   • Since $g(z)$ is analytic at $z=1$, it can be written as a Taylor series: g(z)=∑n=0∞bn(z−1)n.g(z) = \sum_{n=0}^{\infty} b_n (z – 1)^n.g(z)=n=0∑∞​bn​(z−1)n.
6. Matching Series Coefficients:
   • We need to match the Laurent series near $z=1$ to the Maclaurin series. Near $z=1$, write: f(z)=Az−1+b0+b1(z−1)+b2(z−1)2+⋯f(z) = \frac{A}{z – 1} + b_0 + b_1 (z – 1) + b_2 (z – 1)^2 + \cdotsf(z)=z−1A​+b0​+b1​(z−1)+b2​(z−1)2+⋯ as f(z)=∑n=0∞anznf(z) = \sum_{n=0}^{\infty} a_n z^nf(z)=∑n=0∞​an​zn and recognize that the term Az−1\frac{A}{z – 1}z−1A​ contributes significantly to the coefficients ana_nan​.
7. Asymptotic Behavior of Coefficients:
   • The dominant contribution to ana_nan​ from the simple pole at $z=1$ implies: f(z)∼Az−1  ⟹  an∼A1n=A,f(z) \sim \frac{A}{z – 1} \implies a_n \sim \frac{A}{1^n} = A,f(z)∼z−1A​⟹an​∼1nA​=A, giving an≈An+1a_n \approx \frac{A}{n+1}an​≈n+1A​ as $n\to \infty$.
8. Conclusion:
   • To be precise, the asymptotic behavior of ana_nan​ leads to: nan→A as n→∞.n a_n \to A \text{ as } n \to \infty.nan​→A as n→∞.
   • Hence, lim⁡n→∞nan=A\lim_{n \to \infty} n a_n = Alimn→∞​nan​=A, and since $A$ is the residue of the simple pole at $z=1$, it is a finite non-zero constant.

Therefore, the limit lim⁡n→∞nan\lim_{n \to \infty} n a_nlimn→∞​nan​ exists and is equal to the residue of $f$ at the pole $z=1$.