Question 1094327
<br>The infinite sum is
{{{5/(1-1/5) = 5/(4/5) = 25/4 = 6.25}}}<br>
Now look at the sequence consisting of the infinite sum minus the sum of the first n terms of the given sequence.  The sums of the first n terms of the given sequence form the sequence
5, 6, 6.2, 6.24, ...<br>
The sequence consisting of the  differences between the infinite sum and the terms in this sequence is
1.25, .25, .05, .01, ...<br>
We can see that this sum is a decreasing geometric sequence, with first term 1.25 and common ratio 0.2.  We want to know how many terms we need to go out in this sequence to get a number less than 2.5x10^-8.  So we need to solve
{{{1.25(0.2)^(n-1) < 2.5*10^-8}}}<br>
Logarithms and a scientific calculator show that 12 terms is just barely not enough; we need to add 13 terms of the given sequence to get within 2.5*10^-8 of the infinite sum.