Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 32998 by abdo imad last updated on 09/Apr/18

$$calculate\sum _{n=1}^{\mathrm{\infty }}\frac{4n+1}{n^2{(3n+1)}^2}.$$

Commented by abdo imad last updated on 15/Apr/18

$$letdecomposeF\left(x\right)=\frac{4x+1}{x^2{(3x+1)}^2}$$$$F\left(x\right)=\frac{a}{x}+\frac{b}{x^2}+\frac{c}{3x+1}+\frac{d}{{(3x+1)}^2}$$$$b={lim}_{x\to 0}{x}^{2}F\left(x\right)=1$$$$d={lim}_{x\to -\frac{1}{3}}{(3x+1)}^{2}F\left(x\right)=\frac{-\frac{4}{3}+1}{\frac{1}{9}}=9.(-\frac{1}{3})=-3$$$$F\left(x\right)=\frac{a}{x}+\frac{1}{x^2}+\frac{c}{3x+1}-\frac{3}{{(3x+1)}^2}$$$${lim}_{x\to \mathrm{\infty }}xF\left(x\right)=0=a+c\Rightarrow c=-a\Rightarrow$$$$F\left(x\right)=\frac{a}{x}+\frac{1}{x^2}-\frac{a}{3x+1}-\frac{3}{{(3x+1)}^2}$$$$F\left(1\right)=\frac{5}{16}=a+1-\frac{a}{4}-\frac{3}{16}=\frac{3}{4}a+\frac{13}{16}\Rightarrow$$$$5=12a+13\Rightarrow 12a=5-13=-8\Rightarrow a=-\frac{8}{12}=-\frac{2}{3}$$$$F\left(x\right)=\frac{-2}{3x}+\frac{1}{x^2}+\frac{2}{3(3x+1)}-\frac{3}{{(3x+1)}^2}$$$$\sum _{k=1}^n\frac{4k+1}{k^2{(3k+1)}^2}=-\frac{2}{3}\sum _{k=1}^n\frac{1}{k}+\sum _{k=1}^n\frac{1}{k^2}$$$$+\frac{2}{3}\sum _{k=1}^n\frac{1}{3k+1}-3\sum _{k=1}^n\frac{1}{{(3k+1)}^2}$$$$=\frac{2}{3}\sum _{k=1}^n(\frac{1}{3k+1}-\frac{1}{k})+\sum _{k=1}^n\frac{1}{k^2}-3\sum _{k=1}^n\frac{1}{{(3k+1)}^2}$$$$but\sum _{k=1}^n\frac{1}{3k+1}=\frac{1}{4}+\frac{1}{7}+\frac{1}{10}+....+\frac{1}{3n+1}$$$$\sum _{k=1}^n\frac{1}{k}=\sum _{k=3p}(...)+\sum _{k=3p+1}(...)+\sum _{k=3p+2}(...)$$$$=\sum _{p=1}^{\left[\frac{n}{3}\right]}\frac{1}{3p}+\sum _{k=0}^{\left[\frac{n-1}{3}\right]}\frac{1}{3p+1}+\sum _{p=0}^{\left[\frac{n-2}{3}\right]}\frac{1}{3p+2}...becontinued...$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com