Math Contest #24 Results and Solution

Solution

The problem of this contest was to find the limit of a converging sum:

This may look unsolvable at first, but it really isn't that hard. You can get to the result with just a bit of restructuring(Due to the lack of latex support at steemit I will use Σ to denote the sume from 1 to infinity. I also define L as the limit of the sum):
L = Σ 1/(x*(x+3)) = Σ (x+3 + 1 -x-3)/(x*(x+3))
= Σ (x+4)/(x*(x+3)) – Σ (x+3)/(x*(x+3))
= Σ (x+4)/(x*(x+3)) – Σ 1/x
Here it is possible to make an index shift by 3 on the right side(thus removing the first 3 terms from the sum and changing x). This is allowed because we know 1/xis converging to 0:
= Σ (x+4)/(x*(x+3)) – Σ 1/(x+3) – 1 – 1/2 – 1/3
= Σ (x+4)/(x*(x+3)) – Σ x/(x*(x+3)) – 1 – 1/2 – 1/3
= Σ (x+4 – x)/(x*(x+3)) – 3/2 – 1/3
= Σ 4/(x*(x+3)) – 11/6
= 4L – 11/6
↓ Rejoining the beginning and the end of this long set of equations
L = 4L - 11/6
3L = 11/6
L = 11/18

↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓

List of participants with their entries:

Name	solution found	comment
@tonimontana	correct
@crokkon	incorrect(0.611111100938577 with computer program)	inaccurate and not done mathematical.

↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓

Winner draw:

Not necessary since only 1 successful participant:
Congratulations @tonimontana , you won 2 SBI!
↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓

The next contest starts soon. Don't miss it!