Question

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

Answer 1

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.

Answer 2

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.