How do you determine the convergence or divergence of $\infty \sum n = 1 \frac{{(- 1)}^{n + 1}}{n}$ $sum_(n=1)^(oo) (-1)^(n+1)/n$ ?

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Answer 1 · 2017-01-23T14:10:25

The series is convergent and its sum is:

$\infty \sum n = 1 \frac{{(- 1)}^{n + 1}}{n} = ln 2$ $sum_(n=1)^oo (-1)^(n+1)/n = ln2$

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

$\infty \sum n = 1 {(- 1)}^{n} a_{n}$ $sum_(n=1)^oo (-1)^n a_n$
to be convergent is that:

(i) $a_{n + 1} \leq a_{n}$ $a_(n+1) <= a_n$

(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:

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

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