How do you find the sum of the infinite geometric series 1 + 2 + 4 + 8 +… ?

2 Answers
Nov 7, 2015

This series diverges. Its sum is #oo#.

Explanation:

A infinite geometric series of the form #sum_(r=1)^ooar^(n-1)# converges #iff |r|<1#.

In this case the common ratio #r=x_(n+1)/(x_n)=2>1# hence the series diverges.

Nov 7, 2015

The series diverges, that is, it has no finite sum.

Explanation:

For an infinite series to converge to a specific sum, the terms of the series need to approach 0. In this case, not only do the terms not approach 0, but they increase by a factor of 2. It should be intuitive, then, that the series increases without bound, as you are adding an infinite number of increasing values.


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For a slightly more rigorous proof that does not rely on knowledge of series, suppose that there exists a number #n# such that
#1+2+4+8+... = n#

Let #S_k = 1+2+4+...+2^(k-1)# (the sum of the first #k# terms of the series)
and let #N# be an integer greater than #n#.

Note that for any integer #k#
#1+2+4+...+2^(k-1) > 1+1+1+...+1 = k#

Then we have
#n = 1+2+4+8+...>S_N>N>n#

But #n>n# is a contradiction, and so no such #n# exists.