(1)
\[
\sum_{n=1}^{+\infty}\frac{1}{n(2n-1)}
\]
(2)
\[
\sum_{n=1}^{+\infty}(\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n})
\]
(3)
\[
\sum_{n=1}^{+\infty}\frac{1}{n(n+3)}
\]
(4)
\[
\sum_{n=1}^{+\infty}\frac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}
\]
(5)
\[
\sum_{n=1}^{+\infty}\frac{(-1)^{n-1}}{n}
\]
(6)
\[
\sum_{n=1}^{+\infty}\frac{1}{n(n+1)(n+2)}
\]
1
前 $N$ 项和为
\[
\begin{split}
S_N&=\sum_{n=1}^{N}\frac{1}{n(2n-1)}=2\sum_{n=1}^{N}\frac{1}{2n(2n-1)}=2\sum_{n=1}^{N}\Bigl[\frac{1}{2n-1}-\frac{1}{2n}\Bigr]\\
&=2\cdot\Bigl[(1-\frac{1}{2})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{5}-\frac{1}{6})+\cdots+(\frac{1}{2N-1}-\frac{1}{2N})\Bigr]
\end{split}
\]
或者写成下面的形式
\[
\begin{split}
S_N&=\sum_{n=1}^{N}\frac{1}{n(2n-1)}=\sum_{n=1}^{N}\frac{1}{2}\cdot\frac{1}{n(n-\frac{1}{2})}\\
&=\sum_{n=1}^{N}\Bigl[\frac{1}{n-\frac{1}{2}}-\frac{1}{n}\Bigr]\\
&=(\frac{1}{\frac{1}{2}}-\frac{1}{1})+(\frac{1}{\frac{3}{2}}-\frac{1}{2})+(\frac{1}{\frac{5}{2}}-\frac{1}{3})+\cdots+(\frac{1}{\frac{2N-1}{2}}-\frac{1}{N})\\
&=(\frac{2}{1}-1)+(\frac{2}{3}-\frac{1}{2})+(\frac{2}{5}-\frac{1}{3})+\cdots+(\frac{2}{2N-1}-\frac{1}{N})\\
&=2(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots+\frac{1}{2N-1})-(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{N})
\end{split}
\]
注意到
\[
\ln 2=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots+\frac{(-1)^{n-1}}{n}+\cdots
\]
因此, 原级数的和为 $2\ln 2$.
2
(2)
前 $n$ 项和(或第 $n$ 个部分和)为
\[
\begin{split}
S_n&=\sum_{k=1}^{n}(\sqrt{k+2}-2\sqrt{k+1}+\sqrt{k})\\
&=\ \ \ (\sqrt{3}-2\sqrt{2}+\sqrt{1})\\
&\quad+(\sqrt{4}-2\sqrt{3}+\sqrt{2})\\
&\quad+(\sqrt{5}-2\sqrt{4}+\sqrt{3})\\
&\quad+(\sqrt{6}-2\sqrt{5}+\sqrt{4})\\
&\quad+(\sqrt{7}-2\sqrt{6}+\sqrt{5})\\
&\quad+\cdots\\
&\quad+(\sqrt{n-2}-2\sqrt{n-3}+\sqrt{n-4})\\
&\quad+(\sqrt{n-1}-2\sqrt{n-2}+\sqrt{n-3})\\
&\quad+(\sqrt{n}-2\sqrt{n-1}+\sqrt{n-2})\\
&\quad+(\sqrt{n+1}-2\sqrt{n}+\sqrt{n-1})\\
&\quad+(\sqrt{n+2}-2\sqrt{n+1}+\sqrt{n})\\
&=\sqrt{n+2}-\sqrt{n+1}-\sqrt{2}+1
\end{split}
\]
因此,
\[
\begin{split}
\lim_{n\rightarrow\infty}S_n&=\lim_{n\rightarrow\infty}(\sqrt{n+2}-\sqrt{n+1}-\sqrt{2}+1)\\
&=\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n+2}+\sqrt{n+1}}+1-\sqrt{2}\\
&=1-\sqrt{2}.
\end{split}
\]
3
(3)
\[
\sum_{n=1}^{\infty}\frac{1}{n(n+3)}
\]
前 $n$ 项和为
\[
\begin{split}
S_n&=\sum_{k=1}^{n}\frac{1}{k(k+3)}=\sum_{k=1}^{n}\frac{1}{3}(\frac{1}{k}-\frac{1}{k+3})\\
&=\frac{1}{3}\biggl[(1-\frac{1}{4})+(\frac{1}{2}-\frac{1}{5})+(\frac{1}{3}-\frac{1}{6})+(\frac{1}{4}-\frac{1}{7})+(\frac{1}{5}-\frac{1}{8})+\cdots\\
&\qquad+(\frac{1}{n-3}-\frac{1}{n})+(\frac{1}{n-2}-\frac{1}{n+1})+(\frac{1}{n-1}-\frac{1}{n+2})+(\frac{1}{n}-\frac{1}{n+3})\biggr]\\
&=\frac{1}{3}\biggl[1+\frac{1}{2}+\frac{1}{3}-\frac{1}{n+1}-\frac{1}{n+2}-\frac{1}{n+3}\biggr]
\end{split}
\]
因此
\[
\lim_{n\rightarrow\infty}S_n=\frac{1}{3}\biggl[1+\frac{1}{2}+\frac{1}{3}\biggr]=\frac{11}{18}.
\]
即
\[
\sum_{n=1}^{\infty}\frac{1}{n(n+3)}=\frac{11}{18}.
\]
使用 Sowya 进行计算.
>> setprecision(100)
Now the precision is: 100
------------------------
>> sum(1/(n*(n+3)),n,1,1000)
0.6101131064564118695138808451564793317976881060086533486692291714351284144897317355989244515283165815
------------------------
>> sum(1/(n*(n+3)),n,1,10000)
0.6110111311064456441178697513188288391174442574919318512318056899799952708981013687622035058415362162
------------------------
>> :mode fraction
Switch into fraction calculating mode.
e.g., 1/2+1/3 will return 5/6
>> sum(1/(n*(n+3)),n,1,10000)
305688902500|500300055003
------------------------
>> :mode numerical
Switch into numerical calculating mode.
e.g., 1/2+1/3 will return 0.83333333
>> 305688902500/500300055003
in> 305688902500/500300055003
out> 0.6110111311064456441178697513188288391174442574919318512318056899799952708981013687622035058415362147
------------------------
>> 11/18
in> 11/18
out> 0.6111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111111
------------------------