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Assuming that you are referring to the sum of the infinite sequence of a,

 

Let's try to expand the terms

\(\begin{align} a_1 &=( \frac{1}{2}) \\\\ a_2 &= \frac{1}{6} = \frac{3}{6}-\frac{2}{6} \\ &=( \frac{1}{2}) -( \frac{1}{3})\\\\ a_3 &= \frac{1}{12} = \frac{4}{12}-\frac{3}{12} \\ &=( \frac{1}{3}) -( \frac{1}{4}) \\\text{and so on...} \end{align}\)

Hence, we can conclude that every term of a_n (apart from a₁) equals to \(\frac {1}{n} - \frac {1}{n+1} \)

Therefore, the sum of such an infinite sequence is presented as ...\(a_1 + a_2 + a_3 + \ldots =\frac{1}{2} + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{5}\right) + \ldots +\left(\frac{1}{n-2} - \frac{1}{n-1}\right) + \left(\frac{1}{n-1} - \frac{1}{n}\right))\)

... where n in this case would be infinity,

equals to...

\(a_1 + a_2 + a_3 + \ldots =\frac{1}{2} + \frac{1}{2} - \frac{1}{n} = 1 - \frac{1}{n}\)

 

the sum of the sequence approaches 1 

further explanation (to the best of my abilities below)

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As you can imagine, even though \(\frac {1}{n} \) never reaches 0 but continues to decrease in value,

there is a point where the value of \(\frac {1}{n} \) would be so small

 

(think that when n = one trillion, \(\frac {1}{n} \) would result in 0.000000000001

1 - 0.000000000001 = 0.99999999999 

 - and n, as infinity, would not stop at one trillion,

meaning that as the sequence goes on, smaller and smaller values of  \(\frac {1}{n} \) would be subtracted from 1, )

 

 

Hence, we completely ignore the value of n, instead we can just say that 

the sum of the sequence approaches 1 

Using an online  Sigma calculator 

 

 

\(\sum_{1}^{∞}\)1/(n(n+1)    =   1 

Found a solution: 1/n(n+1)=(1/n)-(1/n+1)

So It's (1-1/2)+(1/2-1/3)...

Everything cancels out to one