Question

# Sum the following series to n terms and to infinity $$\displaystyle\frac{1}{1.4.7}+\frac{1}{4.7.10}+\frac{1}{7.10.13}+...$$.

Solution

## $${ n }^{ th }$$ term $$T_{n}=\cfrac { 1 }{ \left( 3n-2 \right) \left( 3n+1 \right) \left( 3n+4 \right) }$$By partial fraction we get$${ T }_{ n }=\cfrac { 1 }{ 6 } \left[ \cfrac { 1 }{ \left( 3n-2 \right) \left( 3n+1 \right) } -\cfrac { 1 }{ \left( 3n+1 \right) \left( 3n+4 \right) } \right]$$$${ T }_{ 1 }=\cfrac { 1 }{ 6 } \left[ \cfrac { 1 }{ 1.4 } -\cfrac { 1 }{ 4.7 } \right]$$$${ T }_{ 2 }=\cfrac { 1 }{ 6 } \left[ \cfrac { 1 }{ 4.7 } -\cfrac { 1 }{ 7.10 } \right]$$$${ T }_{ n }=\cfrac { 1 }{ 6 } \left[ \cfrac { 1 }{ \left( 3n-2 \right) \left( 3n+1 \right) } -\cfrac { 1 }{ \left( 3n+1 \right) \left( 3n+4 \right) } \right]$$Then, $${ S }_{ n }=\displaystyle \sum_{n=1}^{n} T_{n}=\cfrac { 1 }{ 6 } \left[ \cfrac { 1 }{ 1.4 } -\cfrac { 1 }{ \left( 3n+1 \right) \left( 3n+4 \right) } \right]$$$$=\cfrac { 1 }{ 24 } -\cfrac { 1 }{ 6\left( 3n+1 \right) \left( 3n+4 \right) }$$As $$n\rightarrow \infty$$$$S_{n}=\cfrac { 1 }{ 24 } -0$$$$=\cfrac { 1 }{ 24 }$$Mathematics

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