A Series Problem!

Dear friends i am having a problem in series with factorials

∑r=1nr2−r−1(r+1)!\displaystyle \sum _{ r=1 }^{n }{ \frac { { r }^{ 2 }-r-1 }{ (r+1)! } }

i need to know how to solve the above problem. a solution would be welcome.

#Factorials #SumOfSeries

Note by Nishant Singh
5 years, 7 months ago

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Comments

Do it like this:∑r=1nr2−r−1(r+1)!=∑r=1n(r2+r)−2(r+1)+1(r+1)!=∑r=1n{(r2+r)(r+1)!−2(r+1)(r+1)!+1(r+1)!}=∑r=1n(r2+r)(r+1)!−2∑r=1n(r+1)(r+1)!+∑r=1n1(r+1)!=∑r=1n1(r−1)!−2∑r=1n1r!+∑r=1n1(r+1)!=∑x=0n−11x!−2∑x=0n−11(x+1)!+∑x=0n−11(x+2)!=(10!+11!+12!+...+1(n−1)!)−2(11!+12!+...+1n!)+(12!+13!+...+1(n+1)!)=10!−11!−1n!+1(n+1)!=−n(n+1)!\sum _{ r=1 }^{ n }{ \frac { { r }^{ 2 }-r-1 }{ \left( r+1 \right) ! } } \\ =\sum _{ r=1 }^{ n }{ \frac { \left( { r }^{ 2 }+r \right) -2\left( r+1 \right) +1 }{ \left( r+1 \right) ! } } \\ =\sum _{ r=1 }^{ n }{ \left\{ \frac { \left( { r }^{ 2 }+r \right) }{ \left( r+1 \right) ! } -\frac { 2\left( r+1 \right) }{ \left( r+1 \right) ! } +\frac { 1 }{ \left( r+1 \right) ! } \right\} } \\ =\sum _{ r=1 }^{ n }{ \frac { \left( { r }^{ 2 }+r \right) }{ \left( r+1 \right) ! } } -2\sum _{ r=1 }^{ n }{ \frac { \left( r+1 \right) }{ \left( r+1 \right) ! } } +\sum _{ r=1 }^{ n }{ \frac { 1 }{ \left( r+1 \right) ! } } \\ =\sum _{ r=1 }^{ n }{ \frac { 1 }{ \left( r-1 \right) ! } } -2\sum _{ r=1 }^{ n }{ \frac { 1 }{ r! } } +\sum _{ r=1 }^{ n }{ \frac { 1 }{ \left( r+1 \right) ! } } \\ =\sum _{ x=0 }^{ n-1 }{ \frac { 1 }{ x! } } -2\sum _{ x=0 }^{ n-1 }{ \frac { 1 }{ \left( x+1 \right) ! } } +\sum _{ x=0 }^{ n-1 }{ \frac { 1 }{ \left( x+2 \right) ! } }\\=\left( \frac { 1 }{ 0! } +\frac { 1 }{ 1! } +\frac { 1 }{ 2! } +...+\frac { 1 }{ \left( n-1 \right) ! } \right) -2\left( \frac { 1 }{ 1! } +\frac { 1 }{ 2! } +...+\frac { 1 }{ n! } \right) +\left( \frac { 1 }{ 2! } +\frac { 1 }{ 3! } +...+\frac { 1 }{ \left( n+1 \right) ! } \right) \\ =\frac { 1 }{ 0! } -\frac { 1 }{ 1! } -\frac { 1 }{ n! } +\frac { 1 }{ \left( n+1 \right) ! } \\ =-\frac { n }{ \left( n+1 \right) ! }

Kuldeep Guha Mazumder - 5 years, 6 months ago

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Thank You! for the solution

nishant singh - 5 years, 6 months ago

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You're welcome! Hope I could make it clear..

Kuldeep Guha Mazumder - 5 years, 6 months ago

Hint:- Try to find a telescoping series.

Siddhartha Srivastava - 5 years, 7 months ago

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Tried it but I'm unable to split in partial fractions.

nishant singh - 5 years, 7 months ago

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You should be getting partial fractions of the form r(r+1)! \frac{r}{(r+1)!} .

Siddhartha Srivastava - 5 years, 7 months ago
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