$$ { 解：\lim_{n\rightarrow \infty} \left( \frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+\cdots +\frac{2n-1}{2^n} \right) =\lim_{n\rightarrow \infty} \sum_{k=1}^n{\frac{2k-1}{2^k}}=\sum_{n=1}^{\infty}{\frac{2n-1}{2^n}}；} \\ { 设S\left( x \right) =\sum_{n=1}^{\infty}{\frac{2n-1}{2^n}}x^{2n-2}，其收敛域为x\in \left( -\sqrt{2},\sqrt{2} \right) .} \\ { 利用幂级数和函数的逐项求导和逐项积分的方法有：} \\ { S\left( x \right) \begin{array}{l} =\left( \int_0^x{\sum_{n=1}^{\infty}{\frac{2n-1}{2^n}}t^{2n-2}}\mathrm{d}t \right) ^{\prime}=\left( \sum_{n=1}^{\infty}{\frac{2n-1}{2^n}}\int_0^x{t^{2n-2}}\mathrm{d}t \right) ^{\prime}=\left( \sum_{n=1}^{\infty}{\frac{2n-1}{2^n}\cdot \frac{1}{2n-1}}t^{2n-1}\mid_{0}^{x} \right) ^{\prime}\\ =\left( \sum_{n=1}^{\infty}{\frac{x^{2n-1}}{2^n}} \right) ^{\prime}=\left( \frac{1}{x}\sum_{n=1}^{\infty}{\frac{x^{2n}}{2^n}} \right) ^{\prime}=\left[ \frac{1}{x}\sum_{n=1}^{\infty}{\left( \frac{x^2}{2} \right) ^n} \right] ^{\prime}=\left[ \frac{1}{x}\cdot \frac{\frac{1}{2}x^2}{1-\frac{1}{2}x^2} \right] ^{\prime}=\left( \frac{x}{2-x^2} \right) ^{\prime}\\ =\frac{2-x^2-\left[ x\cdot \left( -2x \right) \right]}{\left( 2-x^2 \right) ^2}=\frac{2-x^2+2x^2}{\left( 2-x^2 \right) ^2}=\frac{2+x^2}{\left( 2-x^2 \right) ^2}；\\ \end{array}} \\ { \lim_{n\rightarrow \infty} \left( \frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+\cdots +\frac{2n-1}{2^n} \right) =\sum_{n=1}^{\infty}{\frac{2n-1}{2^n}}=S\left( 1 \right) =\frac{2+1^2}{\left( 2-1^2 \right) ^2}=3.} \\ { 注1：收敛域是\left( -\sqrt{2},\sqrt{2} \right) 的原因} \\ { S\left( \sqrt{2} \right) =\sum_{n=1}^{\infty}{\frac{2n-1}{2^n}}\left( 2^{\frac{1}{2}} \right) ^{2n-2}=\sum_{n=1}^{\infty}{\frac{2n-1}{2^n}}\cdot 2^{n-1}=\sum_{n=1}^{\infty}{\frac{2n-1}{2}}=\sum_{n=1}^{\infty}{\left( n-\frac{1}{2} \right) ，发散；}} \\ { S\left( -\sqrt{2} \right) =\sum_{n=1}^{\infty}{\frac{2n-1}{2^n}}\left( -2^{\frac{1}{2}} \right) ^{2n-2}=\sum_{n=1}^{\infty}{\frac{2n-1}{2^n}}\cdot \left( -2^{n-1} \right) =-\sum_{n=1}^{\infty}{\frac{2n-1}{2}}=-\sum_{n=1}^{\infty}{\left( n-\frac{1}{2} \right) ，发散；}} \\ { 实际上，在\frac{1}{x}\sum_{n=1}^{\infty}{\left( \frac{x^2}{2} \right) ^n}这一步的时候，更能体现出\left( -\sqrt{2},\sqrt{2} \right) 是收敛域.} \\ { 注2：为什么和函数设为：S\left( x \right) =\sum_{n=1}^{\infty}{\frac{2n-1}{2^n}}x^{2n-2}} \\ { 我们注意到2n-1与x^{2n-2}恰好差一位，可以通过逐项积分去掉2n-1，这样让幂级数变得简单.} \\ { 注3：\sum_{n=1}^{\infty}{x^n}=\sum_{n=0}^{\infty}{x^n}-1=\frac{1}{1-x}-1=\frac{1}{1-x}-\frac{1-x}{1-x}=\frac{x}{1-x}，\left| x \right|<1.} \\ { 解法2：通过作差求解} \\ { 设x_n=\frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+\cdots +\frac{2n-1}{2^n}\cdots \cdots \unicode{x2460} ；则2x_n=1+\frac{3}{2}+\frac{5}{2^2}+\cdots +\frac{2n-1}{2^{n-1}}\cdots \cdots \unicode{x2461} ；} \\ { \unicode{x2461} -\unicode{x2460} 有：2x_n-x_n\begin{array}{l} =1+\left( \frac{3}{2}-\frac{1}{2} \right) +\left( \frac{5}{2^2}-\frac{3}{2^2} \right) +\cdots +\left( \frac{2n-1}{2^{n-1}}-\frac{2n-3}{2^{2n-1}} \right) -\frac{2n-1}{2^n}\\ =1+1+\frac{2}{2^2}+\cdots +\frac{2}{2^{n-1}}-\frac{2n-1}{2^n}\\ =1+1+\frac{1}{2^1}+\cdots +\frac{1}{2^{n-2}}-\frac{2n-1}{2^n}\\ =2+\frac{\frac{1}{2}\cdot \left[ 1-\left( \frac{1}{2} \right) ^{n-2} \right]}{1-\frac{1}{2}}-\frac{2n-1}{2^n}\\ =2+\left[ 1-\left( \frac{1}{2} \right) ^{n-2} \right] -\frac{2n-1}{2^n}；\\ \end{array}} \\ { \lim_{n\rightarrow \infty} \left( \frac{1}{2}+\frac{3}{2^2}+\frac{5}{2^3}+\cdots +\frac{2n-1}{2^n} \right) \begin{array}{l} =\lim_{n\rightarrow \infty} x_n=\lim_{n\rightarrow \infty} \left\{ 2+\left[ 1-\left( \frac{1}{2} \right) ^{n-2} \right] -\frac{2n-1}{2^n} \right\}\\ =2+1-0-0=3.\\ \end{array}} \\ { 注：1+\frac{1}{2^1}+\cdots +\frac{1}{2^{n-2}}是等比数列求和，其中a_1=\frac{1}{2}，q=\frac{1}{2}，共n-2项.} \\ $$