Expert: Ahmed Salami - 2/8/2006

Question
Use mathematical induction to prove:
1/3 +1/9 +1/27...+ 1/3^(n)< 1 for n> and =1

Answer
Hi Bryan,
Sorry for the time it took.
The series 1/3 + 1/9 + 1/27 + ..... + 1/(3^n) is a
geometric series with a first term of 1/3 and a common
ratio of 1/3.
By the geometric series formula, the sum to n terms of
the series
S(n) = (1/3)[1 - (1/3)^n]/ (1 - 1/3)
    = (1/3)[1 - 1/(3^n)]/ (2/3)
    = (1/2)[1 - 1/(3^n)]
By inspection, we know that 1/(3^n) is always less than
1 and therefore S(n) is always less than 1/2, which of
course makes it always less than 1. But we've been asked
to prove this by the method of mathematical induction
which is what we do next.
We have to prove that the statement is true for n = 1,
as well as for any n = k+1 whenever it is true for n = k
And so for our problem,
For n = 1,
S(1) = (1/2)[1 - 1/(3^1)]
    = (1/2)[1 - 1/3]
    = (1/2)(2/3)
    = 1/3
i.e S(1) < 1
For n = k,
S(k) = (1/2)[1 - 1/(3^k)]
For n = k+1
S(k+1) = (1/2)[1 - 1/3^(k+1)]
      = (1/2)[1 - 1/3.(3^k)]
      = (1/2)(1/3)[3 - 1/(3^k)]
      = (1/2)(1/3)[2 + (1 - 1/3^k)]
      = (1/2)(1/3)[2 + 2S(k)]
      = (1/3)[1 + S(k)]
Now, if the statement (i.e S(n) < 1) is true for n = k,
we have S(k) < 1
Then,
S(k+1) < (1/3)[1 + 1]
S(k+1) < (1/3)[2]
S(k+1) < 2/3
i.e S(k+1) < 1
Which is our prove of the situation by the method of
mathematical induction, for we have shown that
1/3 + 1/9 + 1/27 + ..... + 1/(3^n) < 1
for n = 1, and also for any n = k+1 as long as it is
true for n = k.

I hope i have helped. You can always get back to me.
Regards.