Inverse Sum

$\begin{aligned} \sum_{n=1}^\infty \frac{1}{\sum_{k=0}^n k} & = \sum_{n=1}^\infty \frac{1}{\frac{1}{2}n(n+1)} \\ & = \sum_{n=1}^\infty \frac{2}{n(n+1)} \\ & = 2 \sum_{n=1}^\infty \left( \frac{1}{n} - \frac{1}{n+1} \right) \\ & = 2 \left( \sum_{n=1}^\infty \frac{1}{n} - \sum_{n=1}^\infty \frac{1}{n+1} \right) \\ & = 2 \left( \sum_{n=1}^\infty \frac{1}{n} - \sum_{n=2}^\infty \frac{1}{n} \right) \\ & = 2 \left(\frac{1}{1} \right) \\ & = \boxed{2} \end{aligned}$

Consider this as the symbol of sigma(£)
£{1÷(1+2+3+4+5+6+7+8+9+10+11+...+n)}
£{1÷n(n+1)÷2}
£{2÷n (n+1)}
2[(1÷1
2)+(1÷2 3)+(1÷3 4)+......]
2[{1-(1÷2)}+(1÷2)-(1÷3)+(1÷3)-(1÷4)......+1÷infinity]
2[1+{1÷infinity}]
And limit of 1÷infinity. Is zero
So 2[1+0]
=2

Why does the question ask for 3 decimal places when the answer is an integer?