# 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/x`

is 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!

*↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓↑↓*

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