The geometric series formula is \(\displaystyle \sum_{i=0}^{\infty} a(r)^i = a\frac{1}{1-r}\)

Your sum is:

\(\displaystyle \sum_{i=1}^{\infty} 2(\frac{1}{5})^i =2\sum_{i=1}^{\infty} (\frac{1}{5})^i \)

Notice the indices go from \(\displaystyle i=1\) to \(\displaystyle \infty.\) We want it to go from \(\displaystyle i=0\) to \(\displaystyle \infty\) so that we can use the geometric series formula.

The \(\displaystyle i=0\) term is \(\displaystyle (\frac{1}{5})^0 = 1.\)

So I both added \(\displaystyle 1\) (by changing the sum from \(\displaystyle [1,\infty]\) to \(\displaystyle [0,\infty]\) ) and subtracted \(\displaystyle 1.\)

\(\displaystyle = 2\left\{[\sum_{i=0}^{\infty} (\frac{1}{5})^i] -1 \right\} \)

Now, apply the formula for the geometric series on \(\displaystyle [\sum_{i=0}^{\infty} (\frac{1}{5})^i]\) where \(\displaystyle r=\frac{1}{5}\) and \(\displaystyle a=1.\)

\(\displaystyle = 2(\frac{1}{1-\frac{1}{5}} -1) = \frac{1}{2}\)