计算 $\sum_{k=1}^{n}\frac{1}{1+2+\cdots+k}$
计算
\[1+\frac{1}{1+2}+\frac{1}{1+2+3}+\cdots+\frac{1}{1+2+3+\cdots+n}\]
并求极限
\[
\lim_{n\rightarrow\infty}\biggl(1+\frac{1}{1+2}+\frac{1}{1+2+3}+\cdots+\frac{1}{1+2+3+\cdots+n}\biggr)
\]
计算
\[1+\frac{1}{1+2}+\frac{1}{1+2+3}+\cdots+\frac{1}{1+2+3+\cdots+n}\]
并求极限
\[
\lim_{n\rightarrow\infty}\biggl(1+\frac{1}{1+2}+\frac{1}{1+2+3}+\cdots+\frac{1}{1+2+3+\cdots+n}\biggr)
\]
1
\[
\frac{1}{1+2+\cdots+k}=\frac{1}{\frac{k(k+1)}{2}}=\frac{2}{k(k+1)}=2(\frac{1}{k}-\frac{1}{k+1}),
\]
因此
\[
\begin{split}
\sum_{k=1}^{n}\frac{1}{1+2+\cdots+k}&=\sum_{k=1}^{n}2(\frac{1}{k}-\frac{1}{k+1})\\
&=2\Bigl[(\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+\cdots+(\frac{1}{n}-\frac{1}{n+1})\Bigr]\\
&=2(1-\frac{1}{n+1}).
\end{split}
\]
故
\[
\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{1}{1+2+\cdots+k}=2.
\]