Complex Analysis: Theory and Applications · Problem 4.4.3

Elementary Functions — Howell & Matthews

Mathematics — Complex Analysis · radius-of-convergence

Problem

Find the radius of convergence of the following. ((a)) g(z)=∑n=0∞(−1)nzn(2n)!g(z) = \displaystyle\sum_{n=0}^{\infty} (-1)^n \frac{z^n}{(2n)!}. ((b)) h(z)=∑n=0∞n!znh(z) = \displaystyle\sum_{n=0}^{\infty} n! z^n. ((c)) f(z)=∑n=0∞(4n22n+1−6n23n+4)nznf(z) = \displaystyle\sum_{n=0}^{\infty} \left(\frac{4n^2}{2n+1} - \frac{6n^2}{3n+4}\right)^n z^n. ((d)) g(z)=∑n=0∞(n!)2(2n)!zng(z) = \displaystyle\sum_{n=0}^{\infty} \frac{(n!)^2}{(2n)!} z^n. ((e)) h(z)=∑n=0∞(2−(−1)n)nznh(z) = \displaystyle\sum_{n=0}^{\infty} (2-(-1)^n)^n z^n. ((f)) f(z)=∑n=0∞n(n−1)zn(3+4i)nf(z) = \displaystyle\sum_{n=0}^{\infty} \frac{n(n-1)z^n}{(3+4i)^n}. ((g)) g(z)=∑n=0∞(3n+74n+2)nzng(z) = \displaystyle\sum_{n=0}^{\infty} \left(\frac{3n+7}{4n+2}\right)^n z^n. ((h)) h(z)=∑n=0∞2n1+3nznh(z) = \displaystyle\sum_{n=0}^{\infty} \frac{2^n}{1+3^n} z^n. ((i)) g(z)=∑n=0∞z2ng(z) = \displaystyle\sum_{n=0}^{\infty} z^{2n}. ((j)) g(z)=∑n=0∞nnn!zng(z) = \displaystyle\sum_{n=0}^{\infty} \frac{n^n}{n!} z^n. Hint: lim⁡n→∞(1+1n)n=e\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n = e.

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