A geometric series is a series of the form
#a+ar+ar^2+ar^3+...#
where #a# is an initial value and #r# is a common ratio between terms.
In general, the sum of a geometric series is
#a + ar + ar^2 + ... + ar^(n-1) = a(1-r^n)/(1-r)# (see below for derivation)
If #|r| < 1# then as #n ->oo, r^n->0# so the formula reduces to
#a + ar + ar^2 + ... = a/(1-r)#
Now, in order to find #r#, it suffices to divide any term in the series after the first by the term prior to it, as
#(ar^n)/(ar^(n-1))=r#
So, picking the second and first terms, for this series we get
#r = (1/5)/1 = 1/5#
As the first term in the series gives us #a=1# we get the final sum
#1/(1-(1/5)) = 1/(4/5) = 5/4#
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What follows is a short derivation for the geometric series formula, and is not required for understanding the above solution.
Let #S_n# be the #n#th partial sum of a geometric series with ratio #r# and first term #a#, that is,
#S_n = a + ar + ar^2 + ... + ar^(n-1)#
#=>rS_n = ar + ar^2 + ... + ar^n#
#=> S_n - rS_n = a - ar^n#
#=> S_n(1-r) = a(1-r^n)#
#=> S_n = a(1-r^n)/(1-r)#
