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A Series Problem!

Dear friends i am having a problem in series with factorials

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

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

Note by Nishant Singh
1 year, 1 month ago

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Do it like this:\[\sum _{ r=1 }^{ n }{ \frac { { r }^{ 2 }-r-1 }{ \left( r+1 \right) ! } } \\ =\sum _{ r=1 }^{ n }{ \frac { \left( { r }^{ 2 }+r \right) -2\left( r+1 \right) +1 }{ \left( r+1 \right) ! } } \\ =\sum _{ r=1 }^{ n }{ \left\{ \frac { \left( { r }^{ 2 }+r \right) }{ \left( r+1 \right) ! } -\frac { 2\left( r+1 \right) }{ \left( r+1 \right) ! } +\frac { 1 }{ \left( r+1 \right) ! } \right\} } \\ =\sum _{ r=1 }^{ n }{ \frac { \left( { r }^{ 2 }+r \right) }{ \left( r+1 \right) ! } } -2\sum _{ r=1 }^{ n }{ \frac { \left( r+1 \right) }{ \left( r+1 \right) ! } } +\sum _{ r=1 }^{ n }{ \frac { 1 }{ \left( r+1 \right) ! } } \\ =\sum _{ r=1 }^{ n }{ \frac { 1 }{ \left( r-1 \right) ! } } -2\sum _{ r=1 }^{ n }{ \frac { 1 }{ r! } } +\sum _{ r=1 }^{ n }{ \frac { 1 }{ \left( r+1 \right) ! } } \\ =\sum _{ x=0 }^{ n-1 }{ \frac { 1 }{ x! } } -2\sum _{ x=0 }^{ n-1 }{ \frac { 1 }{ \left( x+1 \right) ! } } +\sum _{ x=0 }^{ n-1 }{ \frac { 1 }{ \left( x+2 \right) ! } }\\=\left( \frac { 1 }{ 0! } +\frac { 1 }{ 1! } +\frac { 1 }{ 2! } +...+\frac { 1 }{ \left( n-1 \right) ! } \right) -2\left( \frac { 1 }{ 1! } +\frac { 1 }{ 2! } +...+\frac { 1 }{ n! } \right) +\left( \frac { 1 }{ 2! } +\frac { 1 }{ 3! } +...+\frac { 1 }{ \left( n+1 \right) ! } \right) \\ =\frac { 1 }{ 0! } -\frac { 1 }{ 1! } -\frac { 1 }{ n! } +\frac { 1 }{ \left( n+1 \right) ! } \\ =-\frac { n }{ \left( n+1 \right) ! } \] Kuldeep Guha Mazumder · 12 months ago

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@Kuldeep Guha Mazumder Thank You! for the solution Nishant Singh · 11 months, 3 weeks ago

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@Nishant Singh You're welcome! Hope I could make it clear.. Kuldeep Guha Mazumder · 11 months, 3 weeks ago

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Hint:- Try to find a telescoping series. Siddhartha Srivastava · 1 year, 1 month ago

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@Siddhartha Srivastava Tried it but I'm unable to split in partial fractions. Nishant Singh · 1 year, 1 month ago

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@Nishant Singh You should be getting partial fractions of the form \( \frac{r}{(r+1)!} \). Siddhartha Srivastava · 1 year, 1 month ago

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