سری ها

یادآوری می‌کنیم که:

$\begin{array}{l}\sum _{k=1}^n\left(f\left(k\right)-f\left(k+1\right)\right)=f\left(1\right)-f\left(n+1\right)\end{array}$

بنابه قاعده ادغام با فرض زیر داریم:

$f\left(k\right)=\frac{1}{4^k}$

$\begin{array}{l}{S}_n=\sum _{k=1}^n\left(\frac{1}{4^k}-\frac{1}{4^{k+1}}\right)\end{array}$

$\begin{array}{l}\\ \Rightarrow S_n=f\left(1\right)-f\left(n+1\right);f\left(k\right)=\frac{1}{4^k}\end{array}$

$\begin{array}{l}\\ \Rightarrow S_n=\frac{1}{4}-\frac{1}{4^{n+1}}\end{array}$

$\begin{array}{l}\Rightarrow \underset{n\to +\infty }{\lim }{S}_n=\underset{n\to +\infty }{\lim }\left(\frac{1}{4}-\frac{1}{4^{n+1}}\right)\end{array}$

$\begin{array}{l}\\ \Rightarrow \underset{n\to +\infty }{\lim }{S}_n=\frac{1}{4}-\frac{1}{\infty }\end{array}$

$\begin{array}{l}\Rightarrow \underset{n\to +\infty }{\lim }{S}_n=\frac{1}{4}-0\\ \\ \Rightarrow \underset{n\to +\infty }{\lim }{S}_n=\frac{1}{4}\\ \\ \Rightarrow S=\frac{1}{4}\end{array}$

اين سری به عدد $\frac{1}{4}$ همگراست.

ابتدا کسر را به حاصل جمع دو کسر ساده‌تر تجزيه می‌کنيم:

$\begin{array}{l}\frac{1}{\left(n+2\right)\left(n+3\right)}=\frac{A}{n+2}+\frac{B}{n+3}\end{array}$

$\begin{array}{l}\\ \Rightarrow \frac{1}{\left(n+2\right)\left(n+3\right)}=\frac{A\left(n+3\right)+B\left(n+2\right)}{\left(n+2\right)\left(n+3\right)}\end{array}$

$\begin{array}{l}\Rightarrow \frac{1}{\left(n+2\right)\left(n+3\right)}=\frac{\left(A+B\right)n+3A+2B}{\left(n+2\right)\left(n+3\right)};\left\{\begin{array}{l}A+B=0\\ 3A+2B=1\end{array}\right.〉\left\{\begin{array}{l}A=1\\ B=-1\end{array}\right.\end{array}$

$\Rightarrow \frac{1}{\left(n+2\right)\left(n+3\right)}=\frac{1}{n+2}-\frac{1}{n+3}$

$\begin{array}{l}\Rightarrow \sum _{k=1}^{\infty }\frac{1}{\left(n+2\right)\left(n+3\right)}=\sum _{n=1}^{\infty }\left(\frac{1}{n+2}-\frac{1}{n+3}\right)\end{array}$

$\begin{array}{l}\Rightarrow S=\sum _{k=1}^{+\infty }\left[f_1\left(n\right)-f_1\left(n+1\right)\right];\left(1\right)\end{array}$

$\begin{array}{l}\\ \Rightarrow S=f\left(1\right)-\underset{n\to +\infty }{\lim }f\left(n+1\right)\end{array}$

$\begin{array}{l}\Rightarrow S=\frac{1}{3}-\underset{n\to +\infty }{\lim }\frac{1}{n+3}\\ \\ \Rightarrow S=\frac{1}{3}\end{array}$

$\left(1\right)$: بنا به قاعده ادغام و با فرض زیر، داریم:

$f\left(n\right)=\frac{1}{n+2}$

$S=\sum _{n=1}^{+\infty }\left(f\left(n\right)-f\left(n+1\right)\right)=f\left(1\right)-\underset{n\to +\infty }{\lim }f\left(n+1\right)$

اين سری به عدد $\frac{1}{3}$ همگراست. 

$\begin{array}{l}\log \frac{n^2+2n}{{\left(n+1\right)}^2}=\log \frac{n\left(n+2\right)}{{\left(n+1\right)}^2}\end{array}$

$\begin{array}{l}\\ \Rightarrow \log \frac{n^2+2n}{{\left(n+1\right)}^2}=\log \frac{n}{n+1}\times \frac{n+2}{n+1}\end{array}$

$\begin{array}{l}\Rightarrow \log \frac{n^2+2n}{{\left(n+1\right)}^2}=\log \frac{n}{n+1}+\log \frac{n+2}{n+1}\end{array}$

$\begin{array}{l}\\ \Rightarrow \log \frac{n^2+2n}{{\left(n+1\right)}^2}=\log \frac{n}{n+1}-\log \frac{n+1}{n+2}\end{array}$

$\Rightarrow \sum _{n=1}^{+\infty }\log \frac{n^2+2n}{{\left(n+1\right)}^2}=\sum _{k=1}^{+\infty }\left(\log \frac{n}{n+1}-\log \frac{n+1}{n+2}\right)$

$\begin{array}{l}\Rightarrow S=\sum _{k=1}^{+\infty }\left[f_1\left(n\right)-f_1\left(n+1\right)\right];\left(1\right)\end{array}$

$\begin{array}{l}\\ \Rightarrow S=f\left(1\right)-\underset{n\to +\infty }{\lim }f\left(n+1\right)\end{array}$

$\begin{array}{l}\Rightarrow S=\log \frac{1}{2}-\underset{n\to +\infty }{\lim }\log \frac{n+1}{n+2}\\ \\ \Rightarrow S=\log \frac{1}{2}-\log 1\\ \\ \Rightarrow S=\log \frac{1}{2}\end{array}$

$\left(1\right)$: بنا به قاعده ادغام و با فرض $f\left(n\right)=\log \frac{n}{n+1}$ می‌توان نوشت:

$S=\sum _{n=1}^{+\infty }\left(f\left(n\right)-f\left(n+1\right)\right)=f\left(1\right)-\underset{n\to +\infty }{\lim }f\left(n+1\right)$

اين سری به عدد $\log \frac{1}{2}$ همگراست. 

ابتدا کسر را به حاصل جمع دو کسر ساده ‌تر تجزيه می‌کنيم:

$\begin{array}{l}\frac{1}{k\left(k+2\right)}=\frac{A}{k}+\frac{B}{k+2}\\ \\ \Rightarrow \frac{1}{k\left(k+2\right)}=\frac{A\left(k+2\right)+Bk}{k\left(k+2\right)}\end{array}$

$\begin{array}{l}\Rightarrow \frac{1}{k\left(k+2\right)}=\frac{k\left(A+B\right)+2A}{k\left(k+2\right)};\left\{\begin{array}{l}A=\frac{1}{2}\\ B=-\frac{1}{2}\end{array}\right.\end{array}$

$\Rightarrow \frac{1}{k\left(k+2\right)}=\frac{ \frac{1}{2}}{k}+\frac{ -\frac{1}{2}}{k+2}$

$\begin{array}{l}\Rightarrow \frac{1}{k\left(k+2\right)}=\frac{1}{2}\left(\frac{1}{k}-\frac{1}{k+2}\right)\end{array}$

$\begin{array}{l}\\ \Rightarrow \sum _{k=1}^{+\infty }\frac{1}{k\left(k+2\right)}=\frac{1}{2}\sum _{k=1}^{+\infty }\left(\frac{1}{k}-\frac{1}{k+2}\right)\end{array}$

$\begin{array}{l}\Rightarrow \sum _{k=1}^{+\infty }\frac{1}{k\left(k+2\right)}=\frac{1}{2}\sum _{k=1}^{+\infty }\left[\left(\frac{1}{k}-\frac{1}{k+1}\right)+\left(\frac{1}{k+1}-\frac{1}{k+2}\right)\right]\end{array}$

$\Rightarrow \sum _{k=1}^{+\infty }\frac{1}{k\left(k+2\right)}=\frac{1}{2}\sum _{k=1}^{+\infty }\left(\frac{1}{k}-\frac{1}{k+1}\right)+\frac{1}{2}\sum _{k=1}^n\left(\frac{1}{k+1}-\frac{1}{k+2}\right)$

$\begin{array}{l}\Rightarrow \sum _{k=1}^{+\infty }\frac{1}{k\left(k+2\right)}=\frac{1}{2}\sum _{k=1}^{+\infty }\left[f_1\left(k\right)-f_1\left(k+1\right)\right]+\frac{1}{2}\sum _{k=1}^{+\infty }\left[f_2\left(k\right)-f_2\left(k+1\right)\right];\left({\rm I}\right)\end{array}$

$\begin{array}{l}\Rightarrow S=\frac{1}{2}\left[f_1\left(1\right)-\underset{k\to +\infty }{\lim }{f}_1\left(k+1\right)\right]+\frac{1}{2}\left[f_2\left(1\right)-\underset{k\to +\infty }{\lim }{f}_2\left(k+1\right)\right]\end{array}$

$\begin{array}{l}\Rightarrow S=\frac{1}{2}\left(1-\underset{k\to +\infty }{\lim }\frac{1}{k+1}\right)+\frac{1}{2}\left(\frac{1}{2}-\underset{k\to +\infty }{\lim }\frac{1}{k+2}\right)\end{array}$

$\begin{array}{l}\Rightarrow S=\frac{1}{2}\left(1-0\right)+\frac{1}{2}\left(\frac{1}{2}-0\right)\\ \\ \Rightarrow S=\frac{3}{4}\end{array}$

این سری به عدد فوق همگرا است.

$: \left({\rm I}\right)$ بنا به قاعده ادغام و با فرض زیر داریم:

$\begin{array}{l}{f}_1\left(k\right)=\frac{1}{k},f_2\left(k\right)=\frac{1}{k+1}\end{array}$

$S=\sum _{n=1}^{+\infty }\left(f\left(n\right)-f\left(n+1\right)\right)=f\left(1\right)-\underset{n\to +\infty }{\lim }f\left(n+1\right)$

ابتدا کسر را به حاصل جمع دو کسر ساده ‌تر تجزيه می‌کنيم:

$\begin{array}{l}\frac{2}{\left(k+4\right)\left(k+6\right)}=\frac{A}{k+4}+\frac{B}{k+6}\end{array}$

$\begin{array}{l}\\ \Rightarrow \frac{2}{\left(k+4\right)\left(k+6\right)}=\frac{A\left(k+6\right)+B\left(k+4\right)}{\left(k+4\right)\left(k+6\right)}\end{array}$

$\begin{array}{l}\Rightarrow \frac{2}{\left(k+4\right)\left(k+6\right)}=\frac{\left(A+B\right)K+6A+4B}{\left(k+4\right)\left(k+6\right)};\left\{\begin{array}{l}A+B=0\\ 6A+4B=2\end{array}\right.〉\left\{\begin{array}{l}A=1\\ B=-1\end{array}\right.\end{array}$

$\begin{array}{l}\Rightarrow \frac{2}{\left(k+4\right)\left(k+6\right)}=\frac{1}{k+4}-\frac{1}{k+6}\end{array}$

$\Rightarrow \sum _{k=1}^{+\infty }\frac{2}{\left(k+4\right)\left(k+6\right)}=\sum _{k=1}^{+\infty }\left(\frac{1}{k+4}-\frac{1}{k+6}\right)$

$\begin{array}{l}\Rightarrow \sum _{k=1}^{+\infty }\frac{2}{\left(k+4\right)\left(k+6\right)}=\sum _{k=1}^{+\infty }\left[\left(\frac{1}{k+4}-\frac{1}{k+5}\right)+\left(\frac{1}{k+5}-\frac{1}{k+6}\right)\right]\end{array}$

$\begin{array}{l}\Rightarrow \sum _{k=1}^{+\infty }\frac{2}{\left(k+4\right)\left(k+6\right)}=\sum _{k=1}^{+\infty }\left(\frac{1}{k+4}-\frac{1}{k+5}\right)+\sum _{k=1}^{+\infty }\left(\frac{1}{k+5}-\frac{1}{k+6}\right);\left({\rm I}\right)\end{array}$

$\begin{array}{l}\Rightarrow S=\left[f_1\left(1\right)-\underset{k\to +\infty }{\lim }{f}_1\left(k+1\right)\right]+\left[f_2\left(1\right)-\underset{n\to +\infty }{\lim }{f}_2\left(k+1\right)\right]\end{array}$

$\begin{array}{l}\Rightarrow S=\left[\frac{1}{5}-\underset{k\to +\infty }{\lim }\frac{1}{k+5}\right]+\left[\frac{1}{6}-\underset{k\to +\infty }{\lim }\frac{1}{k+6}\right]\end{array}$

$\begin{array}{l}\Rightarrow S=\frac{1}{5}+\frac{1}{6}\\ \\ \Rightarrow S=\frac{11}{30}\end{array}$

این سری به عدد فوق همگرا است.

$: \left({\rm I}\right)$ بنا به قاعده ادغام و با فرض زیر داریم:

$\begin{array}{l}{f}_1\left(k\right)=\frac{1}{k+4},f_2\left(k\right)=\frac{1}{k+5}\end{array}$

$S=\sum _{n=1}^{+\infty }\left(f\left(n\right)-f\left(n+1\right)\right)=f\left(1\right)-\underset{n\to +\infty }{\lim }f\left(n+1\right)$

ابتدا کسر را به حاصل جمع دو کسر ساده ‌تر تجزيه می‌کنيم:

$\begin{array}{l}\frac{1}{k\left(k+3\right)}=\frac{A}{k}+\frac{B}{k+3}\end{array}$

$\begin{array}{l}\Rightarrow \frac{1}{k\left(k+3\right)}=\frac{ \frac{1}{3}}{k}+\frac{ -\frac{1}{3}}{k+3};\left\{\begin{array}{l}A=\frac{1}{3}\\ B=-\frac{1}{3}\end{array}\right.\end{array}$

$\begin{array}{l}\\ \Rightarrow \frac{1}{k\left(k+3\right)}=\frac{1}{3}\left(\frac{1}{k}-\frac{1}{k+3}\right)\end{array}$

$\begin{array}{l}\Rightarrow \sum _{k=1}^{+\infty }\frac{1}{k\left(k+3\right)}=\frac{1}{3}\sum _{k=1}^{+\infty }\left(\frac{1}{k}-\frac{1}{k+3}\right)\end{array}$

$\Rightarrow \sum _{k=1}^{+\infty }\frac{1}{k\left(k+3\right)}=\frac{1}{3}\sum _{k=1}^{+\infty }\left[\left(\frac{1}{k}-\frac{1}{k+1}\right)+\left(\frac{1}{k+1}-\frac{1}{k+2}\right)+\left(\frac{1}{k+2}-\frac{1}{k+3}\right)\right]$

$\begin{array}{l}\Rightarrow \sum _{k=1}^{+\infty }\frac{1}{k\left(k+3\right)}=\frac{1}{3}\left[\sum _{k=1}^{+\infty }\left(\frac{1}{k}-\frac{1}{k+1}\right)+\sum _{k=1}^{+\infty }\left(\frac{1}{k+1}-\frac{1}{k+2}\right)+\sum _{k=1}^{+\infty }\left(\frac{1}{k+2}-\frac{1}{k+3}\right)\right];\left({\rm I}\right)\end{array}$

$\Rightarrow S=\frac{1}{3}\left[f_1\left(1\right)-\underset{k\to +\infty }{\lim }{f}_1\left(k+1\right)\right]+\frac{1}{3}\left[f_2\left(1\right)-\underset{n\to +\infty }{\lim }{f}_2\left(k+1\right)\right]+\frac{1}{3}\left[f_3\left(1\right)-\underset{n\to +\infty }{\lim }{f}_3\left(k+1\right)\right]$

$\begin{array}{l}\Rightarrow S=\frac{1}{3}\left[\left(1-\underset{k\to +\infty }{\lim }\frac{1}{k+1}\right)+\left(\frac{1}{2}-\underset{k\to +\infty }{\lim }\frac{1}{k+2}\right)+\left(\frac{1}{3}-\underset{k\to +\infty }{\lim }\frac{1}{k+3}\right)\right]\end{array}$

$\begin{array}{l}\Rightarrow S=\frac{1}{3}\left[1+\frac{1}{2}+\frac{1}{3}\right]\\ \\ \Rightarrow S=\frac{11}{18}\end{array}$

این سری به عدد فوق همگرا است.

$: \left({\rm I}\right)$ بنا به قاعده ادغام و با فرض زیر داریم:

$\begin{array}{l}{f}_1\left(k\right)=\frac{1}{k},f_2\left(k\right)=\frac{1}{k+1},f_3\left(k\right)=\frac{1}{k+2}\end{array}$

$S=\sum _{n=1}^{+\infty }\left(f\left(n\right)-f\left(n+1\right)\right)=f\left(1\right)-\underset{n\to +\infty }{\lim }f\left(n+1\right)$

ابتدا کسر را به حاصل جمع دو کسر ساده ‌تر تجزيه می‌کنيم:

$\begin{array}{l}\frac{1}{\left(k+1\right)\left(k+2\right)}=\frac{1}{k+1}-\frac{1}{k+2}\end{array}$

$\begin{array}{l}\Rightarrow \sin \frac{1}{\left(k+1\right)\left(k+2\right)}=\sin \left(\frac{1}{k+1}-\frac{1}{k+2}\right)\end{array}$

$\begin{array}{l}\Rightarrow \sin \frac{1}{\left(k+1\right)\left(k+2\right)}=\sin \frac{1}{k+1}.\cos \frac{1}{k+2}-\cos \frac{1}{k+1}.\sin \frac{1}{k+2}\end{array}$

$\begin{array}{l}\Rightarrow \frac{ \sin \frac{1}{\left(k+1\right)\left(k+2\right)}}{ cos\frac{1}{k+1}.cos\frac{1}{k+2}}=\frac{ \sin \frac{1}{k+1}.\cos \frac{1}{k+2}-\cos \frac{1}{k+1}.\sin \frac{1}{k+2}}{ cos\frac{1}{k+1}.cos\frac{1}{k+2}}\end{array}$

$\Rightarrow \sum _{k=1}^{+\infty }\frac{ \sin \frac{1}{\left(k+1\right)\left(k+2\right)}}{ \cos \frac{1}{k+1}.\cos \frac{1}{k+2}}=\sum _{k=1}^{+\infty }\frac{ \sin \frac{1}{k+1}.\cos \frac{1}{k+2}-\cos \frac{1}{k+1}.\sin \frac{1}{k+2}}{ \cos \frac{1}{k+1}.\cos \frac{1}{k+2}}$

$\begin{array}{l}\Rightarrow \sum _{k=1}^{+\infty }\frac{ \sin \frac{1}{\left(k+1\right)\left(k+2\right)}}{ \cos \frac{1}{k+1}.\cos \frac{1}{k+2}}=\sum _{k=1}^{+\infty }\left(\tan \frac{1}{k+1}-\tan \frac{1}{k+2}\right);\left({\rm I}\right)\end{array}$

$\begin{array}{l}\Rightarrow S=f\left(1\right)-\underset{k\to +\infty }{\lim }f\left(k+1\right)\\ \\ \Rightarrow S=\tan \frac{1}{2}-\underset{k\to +\infty }{\lim }\tan \frac{1}{k+2}\end{array}$

$\begin{array}{l}\Rightarrow S=\tan \frac{1}{2}-\tan 0\\ \\ \Rightarrow S=\tan \frac{1}{2}\end{array}$

این سری به عدد فوق همگرا است.

$: \left({\rm I}\right)$ بنا به قاعده ادغام و با فرض زیر داریم:

$\begin{array}{l}f\left(k\right)=\frac{1}{k+1}\end{array}$

$S=\sum _{n=1}^{+\infty }\left(f\left(n\right)-f\left(n+1\right)\right)=f\left(1\right)-\underset{n\to +\infty }{\lim }f\left(n+1\right)$

یادآوری)

$\sin \left(a-b\right)=\sin a.\cos b-\cos a.\sin b$

ابتدا کسر را به حاصل جمع دو کسر ساده ‌تر تجزيه می‌کنيم:

$\begin{array}{l}\frac{1}{4k^2-1}=\frac{1}{\left(2k-1\right)\left(2k+1\right)}\end{array}$

$\begin{array}{l}\Rightarrow \frac{1}{4k^2-1}=\frac{A}{2k-1}+\frac{B}{2k+1};\left\{\begin{array}{l}A=\frac{1}{2}\\ B=-\frac{1}{2}\end{array}\right.\end{array}$

$\begin{array}{l}\Rightarrow \frac{1}{4k^2-1}=\frac{1}{2}\left(\frac{1}{2k-1}-\frac{1}{2k+1}\right)\end{array}$

$\Rightarrow \sum _{k=1}^n\frac{1}{4k^2-1}=\frac{1}{2}\sum _{k=1}^n\left(\frac{1}{2k-1}-\frac{1}{2k+1}\right);f\left(k\right)=\frac{1}{2k-1}$

$\begin{array}{l}\Rightarrow S=\frac{1}{2}\left[f\left(1\right)-\underset{n\to +\infty }{\lim }f\left(n+1\right)\right]\end{array}$

$\begin{array}{l}\Rightarrow S=\frac{1}{2}\left[\frac{1}{2\left(1\right)-1}-\underset{n\to +\infty }{\lim }\frac{1}{2\left(n+1\right)-1}\right]\end{array}$

$\begin{array}{l}\Rightarrow S=\frac{1}{2}\left(1-0\right)\\ \\ \Rightarrow S=\frac{1}{2}\end{array}$

$=\sum _{n=1}^{\infty }\frac{1}{\sqrt{n+1}+\sqrt{n}}\times \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}-\sqrt{n}}$

$\begin{array}{l}=\sum _{n=1}^{\infty }\left(\sqrt{n+1}-\sqrt{n}\right)\\ \\ =-\sum _{n=1}^{\infty }\left(\sqrt{n}-\sqrt{n+1}\right)\end{array}$

$\begin{array}{l}=-\left[f\left(1\right)-\underset{n\to +\infty }{\lim }f\left(n+1\right)\right]\\ \\ =-\left(1-\underset{n\to +\infty }{\lim }\sqrt{n+1}\right)\end{array}$

$\begin{array}{l}=-1+\underset{n\to +\infty }{\lim }\sqrt{n+1};n\to +\infty \end{array}$

$\begin{array}{l}\\ =+\infty \end{array}$

$\begin{array}{l}=\sum _{n=1}^{+\infty }\left[Ln\left(n\right)-Ln\left(n+1\right)\right];f\left(n\right)=Ln\left(n\right)\end{array}$

$\begin{array}{l}=f\left(1\right)-\underset{n\to +\infty }{\lim }f\left(n+1\right)\\ \\ =Ln1-\underset{n\to +\infty }{\lim }Ln\left(n+1\right)\end{array}$

$\begin{array}{l}=-\underset{n\to +\infty }{\lim }Ln\left(n+1\right)\\ \\ =0-\left(+\infty \right)\\ \\ =-\infty \end{array}$

سری فوق واگرا است.

$\begin{array}{l}=\sum _{n=1}^{+\infty }\left(\frac{1}{5n+2}-\frac{1}{5n+7}\right);f\left(n\right)=\frac{1}{5n+2}\end{array}$

$\begin{array}{l}=f\left(1\right)-\underset{n\to +\infty }{\lim }f\left(n+1\right)\\ \\ =\frac{1}{7}-\underset{n\to +\infty }{\lim }\frac{1}{5n+7}\\ \\ =\frac{1}{7}\end{array}$

$\begin{array}{l}=\sum _{i=1}^n\frac{{\left(i+1\right)}^2-i^2}{i^2{\left(i+1\right)}^2}\end{array}$

$\begin{array}{l}\\ =\sum _{i=1}^n\left[\frac{1}{i^2}-\frac{1}{{\left(i+1\right)}^2}\right];f\left(i\right)=\frac{1}{i^2}\end{array}$

$\begin{array}{l}=f\left(1\right)-f\left(n+1\right)\\ \\ =1-\frac{1}{{\left(n+1\right)}^2}\end{array}$

$\begin{array}{l}=\sum _{k=1}^{+\infty }\frac{2k+3}{\left[{\left(k+1\right)}^2-1\right]\left[{\left(k+2\right)}^2-1\right]}\end{array}$

$\begin{array}{l}=\sum _{k=1}^{+\infty }\left[\frac{1}{{\left(k+1\right)}^2-1}-\frac{1}{{\left(k+2\right)}^2-1}\right];f\left(k\right)=\frac{1}{{\left(k+1\right)}^2-1}\end{array}$

$\begin{array}{l}=f\left(1\right)-\underset{n\to +\infty }{\lim }f\left(k+1\right)\\ \\ =\frac{1}{3}-\underset{n\to +\infty }{\lim }\frac{1}{{\left(k+2\right)}^2-1}\\ \\ =\frac{1}{3}\end{array}$

$\begin{array}{l}=\sum _{n=1}^{+\infty }\frac{1}{\left(3n-1\right)\left(3n+2\right)}\end{array}$

$\begin{array}{l}\\ =\sum _{n=1}^{+\infty }\left[\frac{1}{3\left(3n-1\right)}-\frac{1}{3\left(3n+2\right)}\right]\end{array}$

$\begin{array}{l}=\frac{1}{3}\sum _{n=1}^{+\infty }\left[\frac{1}{\left(3n-1\right)}-\frac{1}{\left(3n+2\right)}\right];f\left(n\right)=\frac{1}{\left(3n-1\right)}\end{array}$

$\begin{array}{l}=\frac{1}{3}\left[f\left(1\right)-\underset{n\to +\infty }{\lim }f\left(n+1\right)\right]\\ \\ =\frac{1}{3}\left[\frac{1}{2}-\underset{n\to +\infty }{\lim }\frac{1}{\left(3n+2\right)}\right]\\ \\ =\frac{1}{6}\end{array}$

$\begin{array}{l}=\frac{2}{2\left(2+1\right)}+\frac{2}{3\left(3+1\right)}+\frac{2}{4\left(4+1\right)}+\mathrm{...}+\frac{2}{n\left(n+1\right)}+\mathrm{...}\end{array}$

$\begin{array}{l}=\sum _{n=2}^{+\infty }\frac{2}{n\left(n+1\right)}\\ \\ =2\sum _{n=2}^{\infty }\frac{1}{n\left(n+1\right)}\end{array}$

$\begin{array}{l}=2\sum _{n=2}^{\infty }\left(\frac{1}{n}-\frac{1}{n+1}\right);f\left(n\right)=\frac{1}{n}\end{array}$

$\begin{array}{l}\\ =2\left[f\left(2\right)-\underset{n\to +\infty }{\lim }f\left(n+1\right)\right]\end{array}$

$\begin{array}{l}=2\left(\frac{1}{2}-\underset{n\to +\infty }{\lim }\frac{1}{n+1}\right)\\ \\ =1\end{array}$

$\begin{array}{l}=\sum _{k=2}^{+\infty }\log \left(\frac{k^2-1}{k^2}\right)\\ \\ =\sum _{k=2}^{+\infty }\log \left(\frac{k-1}{k}.\frac{k+1}{k}\right)\end{array}$

$\begin{array}{l}=\sum _{k=2}^{+\infty }\log \frac{k-1}{k}+\sum _{k=2}^{+\infty }\log \frac{k+1}{k}\end{array}$

$\begin{array}{l}=\sum _{k=2}^{+\infty }\left[\log \left(k-1\right)-\log k\right]+\sum _{k=2}^{+\infty }\left[\log \left(k+1\right)-\log k\right]\end{array}$

$=\sum _{k=2}^{+\infty }\left[\log \left(k-1\right)-\log k\right]-\sum _{k=2}^{+\infty }\left[\log k-\log \left(k+1\right)\right];\left\{\begin{array}{l}{f}_1\left(k\right)=\log \left(k-1\right)\\ f_2\left(k\right)=\log k\end{array}\right.$

$\begin{array}{l}=\left[f_1\left(2\right)-\underset{k\to +\infty }{\lim }{f}_1\left(k+1\right)\right]-\left[f_2\left(2\right)-\underset{k\to +\infty }{\lim }{f}_2\left(k+1\right)\right]\end{array}$

$\begin{array}{l}=\left[\log 1-\underset{k\to +\infty }{\lim }\log k\right]-\left[\log 2-\underset{k\to +\infty }{\lim }{\log }_2\left(k+1\right)\right]\end{array}$

$\begin{array}{l}=-\underset{k\to +\infty }{\lim }\log k-\log 2+\underset{k\to +\infty }{\lim }{\log }_2\left(k+1\right)\end{array}$

$\begin{array}{l}=-\log 2+\underset{k\to +\infty }{\lim }{\log }_2\left(k+1\right)-\underset{k\to +\infty }{\lim }\log k\end{array}$

$\begin{array}{l}=-\log 2+\underset{k\to +\infty }{\lim }\log \frac{\left(k+1\right)}{k}\\ \\ =-\log 2+\underset{k\to +\infty }{\lim }\log 1\end{array}$

$\begin{array}{l}=-\log 2+0\\ \\ =-\log 2\end{array}$

$\begin{array}{l}\sum _{n=1}^{\infty }\frac{n-1}{n!}\end{array}$

$\begin{array}{l}=\sum _{n=1}^{\infty }\left[\frac{1}{\left(n-1\right)!}-\frac{1}{n!}\right];f\left(n\right)=\frac{1}{\left(n-1\right)!}\end{array}$

$\begin{array}{l}=f\left(1\right)-\underset{n\to +\infty }{\lim }f\left(n+1\right)\\ \\ =\frac{1}{0!}-\underset{n\to +\infty }{\lim }\frac{1}{n!}\end{array}$

$\begin{array}{l}=\frac{1}{1}-0\\ \\ =1\end{array}$

$\begin{array}{l}\sum _{k=1}^{\infty }\frac{1}{k^2+3k}\\ \\ =\sum _{k=1}^{\infty }\frac{1}{k\left(k+3\right)};m=3\end{array}$

$\begin{array}{l}=\frac{1}{3}\left(1+\frac{1}{2}+\frac{1}{3}\right)\\ \\ =\frac{1}{3}\left(\frac{6+3+2}{6}\right)\\ \\ =\frac{11}{18}\end{array}$

$\begin{array}{l}\sum _{k=1}^{\infty }\frac{1}{k\left(k+1\right)\left(k+2\right)\left(k+3\right)}\\ \\ =\frac{1}{3}.\frac{1}{3!};m=3\end{array}$

$\begin{array}{l}=\frac{1}{3}\times \frac{1}{6}\\ \\ =\frac{1}{18}\end{array}$