1.确定下列级数的收敛半径：(1)∑k=0∞zkk!解：R=limk→∞∣akak+1∣=limk→∞1k!⋅(k+1)!1=limk→∞(k+1)=∞\begin{aligned} &1.确定下列级数的收敛半径：\\ &(1) \boldsymbol{\sum_{k=0}^{\infty} \frac{z^k}{k!}}\\ &解：R=\lim\limits_{k \to \infty} \left| \frac{a_k}{a_{k+1}} \right| = \lim\limits_{k \to \infty}\frac{1}{k!} \cdot \frac{(k+1)!}{1} = \lim\limits_{k \to \infty} (k+1) = \infty\\ \end{aligned}1.确定下列级数的收敛半径：(1)k=0∑∞k!zk解：R=k→∞lim ak+1ak =k→∞limk!1⋅1(k+1)!=k→∞lim(k+1)=∞
(2)∑k=1∞k2kzk解：R=limk→∞∣akak+1∣=limk→∞k2k⋅2k+1k+1=2⋅limk→∞(kk+1)=2R=limk→∞1∣ak∣k=limk→∞1k2kk=limk→∞2kk=2\begin{aligned} & (2) \boldsymbol{\sum_{k=1}^{\infty} \frac{k}{2^k} z^k}\\ &解：R=\lim\limits_{k \to \infty} \left| \frac{a_k}{a_{k+1}} \right| = \lim\limits_{k \to \infty} \frac{k}{2^k} \cdot \frac{2^{k+1}}{k+1} = 2 \cdot \lim\limits_{k \to \infty} \left( \frac{k}{k+1} \right) = 2\\ &R=\lim\limits_{k \to \infty} \frac{1}{\sqrt[k]{|a_k|}} = \lim\limits_{k \to \infty} \frac{1}{\sqrt[k]{\frac{k}{2^k}}} = \lim\limits_{k \to \infty} \frac{2}{\sqrt[k]{k}} = 2\\ \end{aligned}(2)k=1∑∞2kkzk解：R=k→∞lim ak+1ak =k→∞lim2kk⋅k+12k+1=2⋅k→∞lim(k+1k)=2R=k→∞limk∣ak∣ 1=k→∞limk2kk 1=k→∞limkk 2=2
(3)∑k=1∞k!kkzk解：R=limk→∞∣akak+1∣=limk→∞k!kk⋅(k+1)k+1(k+1)!=limk→∞(k+1k)k(k+1)k+1=e\begin{aligned} & (3) \boldsymbol{\sum_{k=1}^{\infty} \frac{k!}{k^k} z^k}\\ &解：R=\lim\limits_{k \to \infty} \left| \frac{a_k}{a_{k+1}} \right| =\lim\limits_{k \to \infty} \frac{k!}{k^k} \cdot \frac{(k+1)^{k+1}}{(k+1)!} = \lim\limits_{k \to \infty} \left( \frac{k+1}{k} \right)^k \frac{(k+1)}{k+1} = e\\ \end{aligned}(3)k=1∑∞kkk!zk解：R=k→∞lim ak+1ak =k→∞limkkk!⋅(k+1)!(k+1)k+1=k→∞lim(kk+1)kk+1(k+1)=e
(4)∑k=0∞(k+ak)zk解：原级数=∑k=0∞kzk+∑k=0∞akzkR1=limk→∞kk+1=1，R2=1limk→∞∣ak∣k=1∣a∣取小者：{∣a∣≤1, R=1∣a∣>1, R=1∣a∣\begin{aligned} & (4) \boldsymbol{\sum_{k=0}^{\infty} (k+a^k) z^k}\\ &解：原级数=\sum\limits_{k=0}^{\infty} k z^k + \sum\limits_{k=0}^{\infty} a^k z^k\\ &R_1=\lim\limits_{k \to \infty} \frac{k}{k+1}=1，R_2=\frac{1}{\lim\limits_{k \to \infty} \sqrt[k]{|a^k|}}=\frac{1}{|a|}\\ &取小者：\begin{cases} |a| \le 1,\ R=1 \\ |a|>1,\ R=\frac{1}{|a|} \end{cases}\\ \end{aligned}(4)k=0∑∞(k+ak)zk解：原级数=k=0∑∞kzk+k=0∑∞akzkR1=k→∞limk+1k=1，R2=k→∞limk∣ak∣ 1=∣a∣1取小者：{∣a∣≤1, R=1∣a∣>1, R=∣a∣1
(5)∑k=1∞[2+(−1)k]kzk解：用根式判别法R=limk→∞1∣ak∣k=1limk→∞∣[2+(−1)k]k∣k=12+(−1)k{k奇：R=1k偶：R=13，取R=13z=13时，∑k=1∞[2+(−1)k]k(13)k=∑[2+(−1)k3]k，k偶时=1不满足收敛条件。\begin{aligned} & (5) \boldsymbol{\sum_{k=1}^{\infty} [2+(-1)^k]^k z^k}\\ &解：用根式判别法\\ &R=\lim\limits_{k \to \infty} \frac{1}{\sqrt[k]{|a_k|}} =\frac{1}{\lim\limits_{k \to \infty} \sqrt[k]{| [2+(-1)^k]^k |}} = \frac{1}{2+(-1)^k}\\ &\begin{cases} k\text{奇：}R=1 \\ k\text{偶：}R=\frac{1}{3} \end{cases}，取R=\frac{1}{3}\\ &z=\frac{1}{3}时，\sum\limits_{k=1}^{\infty} [2+(-1)^k]^k \left( \frac{1}{3} \right)^k = \sum \left[ \frac{2+(-1)^k}{3} \right]^k，k偶时=1 不满足收敛条件。 \end{aligned}(5)k=1∑∞[2+(−1)k]kzk解：用根式判别法R=k→∞limk∣ak∣ 1=k→∞limk∣[2+(−1)k]k∣ 1=2+(−1)k1{k奇：R=1k偶：R=31，取R=31z=31时，k=1∑∞[2+(−1)k]k(31)k=∑[32+(−1)k]k，k偶时=1不满足收敛条件。