问题及解答

求极限

\[
\lim_{n\rightarrow\infty}\frac{1}{n}(\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\cdots+\frac{n}{2^n})
\]

令

\[
S_n=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\cdots+\frac{n}{2^n},
\]

则

\[
2S_n=1+\frac{2}{2}+\frac{3}{2^2}+\frac{4}{2^3}+\cdots+\frac{n}{2^{n-1}},
\]

两式相减, 得

\[
\begin{split}
S_n&=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\cdots+\frac{1}{2^{n-1}}-\frac{n}{2^n}\\
&=\frac{1-(\frac{1}{2})^n}{1-\frac{1}{2}}-\frac{n}{2^n}.
\end{split}
\]

于是

\[
\lim_{n\rightarrow\infty}\frac{1}{n}S_n=\lim_{n\rightarrow\infty}\frac{1}{n}\biggl[2(1-(\frac{1}{2})^n)-\frac{n}{2^n}\biggr]=0.
\]