Question

How do you determine the convergence or divergence of `sum_(n=1)^(oo) (-1)^(n+1)/n`?

Answer 1

The series is convergent and its sum is:

`sum_(n=1)^oo (-1)^(n+1)/n = ln2`

Explanation:

Leibniz's theorem states that a sufficient condition for the series with alternating signs:

`sum_(n=1)^oo (-1)^n a_n`
to be convergent is that:

(i) `a_(n+1) <= a_n`

(ii) `lim_(n->oo) a_n = 0`

Given: `a_n = 1/n` both conditions are satisfied so the series is convergent.

We can actually calculate its sum starting from the MacLaurin development of the function `ln(1+x)`

In fact:

`ln(1+x)|_(x=0) = 0`

`d/(dx) (ln(1+x)) = 1/(x+1) =(x+1)^(-1)`

`d^2/(dx^2) (ln(1+x)) = -(x+1)^(-2)`

`d^3/(dx^3) (ln(1+x)) = 2(x+1)^-3`

and we can easily conclude that:

`d^n/(dx^n) (ln(1+x)) =( -1)^(n-1)(n-1)! (x+1)^(-n)`

and for `x=0`

`d^n/(dx^n) (ln(1+x))|_(x=0) = ( -1)^(n-1)(n-1)! (1^(-n)) = ( -1)^(n-1)(n-1)!`

so that the MacLaurin series is:

`ln(1+x) = sum_(n=1)^oo (-1)^(n-1)(n-1)! x^n/(n!) = sum_(n=1)^oo (-1)^(n-1) x^n/n`

As `(-1)^(n-1) = (-1)^(n+1)` if we substitute `x=1` we have:

`ln2 = sum_(n=1)^oo (-1)^(n+1)/n`