That's too much Gamma

Algebra Level 3

$\large \sum_{n=1}^{1729} \dfrac{\Gamma(n) \times \Gamma(n+1)}{\Gamma(n+1) \times \Gamma(n+2) }$

If the summation above is the form of $\dfrac ab$ , where $a$ and $b$ are coprime positive integers, find $a+b$ .

Notation : $\Gamma(\cdot)$ denotes the Gamma function .

The answer is 3459.

1 solution

Swapnil Das
Mar 5, 2016

$\large\sum _{ n=1 }^{ 1729 }{ \frac { \Gamma (n)\times \Gamma (n+1) }{ \Gamma (n+1)\times \Gamma (n+2) } }$ = $\large\sum _{ n=1 }^{ 1729 }{ \frac { \Gamma (n) }{ \Gamma (n+2) } } =\sum _{ n=1 }^{ 1729 }{ \frac { \left( n-1 \right) ! }{ \left( n+1 \right) ! } } =\sum _{ n=1 }^{ 1729 }{ \frac { 1 }{ n(n+1) } } =\sum _{ n=1 }^{ 1729 }{ \left( \frac { 1 }{ n } -\frac { 1 }{ n+1 } \right) } =1-\frac { 1 }{ 1730 } =\frac { 1729 }{ 1730 }$

Therefore, $a+b=3459$ .

It looked really scary at first, then I looked up the definition of the gamma function, then the answer instantaneously came to me. Nice problem, though.

Manuel Kahayon - 5 years, 3 months ago

Thanks :) Yes, It does look scary at first :P

Mehul Arora - 5 years, 3 months ago

That's too much Gamma

The answer is 3459.

1 solution

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