Question

What is the sum of the series `1-1+1-1+1-1+...` ?

Answer 1

This series does not converge. It alternates between `0` and `1`.

Explanation:

This is a non-convergent series, which alternates between the values `0` and `1`. It can be thought of as a geometric series with common ratio `-1`.

The general term of a geometric series can be written:

`a_n = a r^(n-1)`

where `a` is the initial term and `r` the common ratio.

Then concerning the partial sum to `N` terms we find:

`(1-r) sum_(n=1)^N a_n = sum_(n=1)^N a r^(n-1) - r sum_(n=1)^N a r^(n-1)`

`color(white)((1-r) sum_(n=1)^N a_n) = sum_(n=1)^N a r^(n-1) - sum_(n=2)^(N+1) a r^(n-1)`

`color(white)((1-r) sum_(n=1)^N a_n) = a + color(red)(cancel(color(black)(sum_(n=2)^N a r^(n-1)))) - color(red)(cancel(color(black)(sum_(n=2)^N a r^(n-1)))) - a r^N`

`color(white)((1-r) sum_(n=1)^N a_n) = a(1 - r^N)`

Dividing both ends by `(1-r)` we get the useful formula:

`sum_(n=1)^N a_n = (a(1-r^N))/(1-r)`

If it converges then:

`sum_(n=1)^oo a_n = lim_(N->oo) sum_(n=1)^N a_n = lim_(N->oo) (a(1-r^N))/(1-r)`

If `abs(r) < 1` then `lim_(N->oo) r^N = 0` and the sum converges to `a/(1-r)`.

If `abs(r) > 1` then `lim_(N->oo) abs(r)^N = oo` and the sum diverges (unless `a = 0`).

If `r = -1` then `(1-r^N)` alternates between `0` and `2`, so does not converge.

Interestingly, if you ignore the fact that `(1-r^N)` does not converge, the formula `a/(1-r)` gives you the value `1/2` as the value of the series, which seems a reasonable compromise, but the partial sums do not converge to it.