Задача 59018 Найти сумму ряда...
Условие
Решение
[m]\frac{1-n}{n(n+1)(n+3)}=\frac{A}{n}+\frac{B}{n+1}+\frac{C}{n+3}[/m]
[m]1-n=A(n+1)(n+3)+Bn(n+3)+Cn(n+1)[/m]
[m]0\cdot n^2-1\cdot n+1=(A+B+C)\cdot n^2+(4A+3B+C)+3A[/m] ⇒
{A+B+C=0 ⇒
{4A+3B+C=-1⇒
{3A=1⇒ А=1/3
{B+C=-1/3 ⇒
{3B+C=-7/3⇒
Вычитаем из второго первое: 2B=-2
B=-1
C=2/3
Находим n-ую частичную сумму
S_(n)=∑ ^(k=n)_(k=1)[m]\frac{1-k}{k(k+1)(k+3)}[/m]=∑ ^(k=n)_(k=1)[m](\frac{1}{3k}-\frac{1}{k+1})+\frac{2}{3(k+3)}=[/m]
=[m](\frac{1}{3}-\frac{1}{2}+\frac{2}{12})+(\frac{1}{6}-\frac{1}{3}+\frac{2}{15})+(\frac{1}{9}-\frac{1}{4}+\frac{2}{18})+[/m]
[m]+(\frac{1}{12}-\frac{1}{5}+\frac{2}{21})...+(\frac{1}{3n}-\frac{1}{n+1}+\frac{2}{3(n+3)})[/m]
Так как [m]\frac{2}{12}+\frac{1}{12}=\frac{3}{12}=\frac{1}{4}[/m]
[m]\frac{1}{4}+(-\frac{1}{4})=0[/m]
[m]\frac{2}{15}+\frac{1}{15}=\frac{3}{15}=\frac{1}{5}[/m]
[m]\frac{1}{5}+(-\frac{1}{5})=0[/m]
...
S_(n)=[m]\frac{1}{3}-\frac{1}{2}+\frac{1}{6}-\frac{1}{3}+\frac{1}{9}+(\frac{2}{3(n+1)})+(\frac{2}{3(n+2)})+(-\frac{1}{n+1}+\frac{2}{3(n+3)})[/m]
S=lim_(n → ∞ )S_(n)=[m]\frac{1}{3}-\frac{1}{2}+\frac{1}{6}-\frac{1}{3}+\frac{1}{9}=-\frac{2}{9}[/m]