# How do you find the nth partial sum, determine whether the series converges and find the sum when it exists given 1-2+3-4+...+n(-1)^(n-1)?

Jan 4, 2018

${\sum}_{n = 1}^{\infty} n {\left(- 1\right)}^{n - 1}$

is indeterminate.

#### Explanation:

The series does not converge as it does not satisfy the necessary condition for convergence:

${\lim}_{n \to \infty} {a}_{n} = 0$

We can evaluate the partial sums noting that for $n$ even we can group the terms as:

${s}_{2 n} = \left(1 - 2\right) + \left(3 - 4\right) + \ldots \left(2 n - 1 - 2 n\right) = {\underbrace{- 1 - 1 + \ldots - 1}}_{n \text{ times}} = - n$

while for $n$ odd:

${s}_{2 n + 1} = 1 + \left(- 2 + 3\right) + \left(- 4 + 5\right) + \ldots + \left(- 2 n + 2 n + 1\right) = {\underbrace{1 + 1 + 1 + \ldots + 1}}_{n + 1 \text{ times}} = n + 1$

Hence the sums oscillate and are unbounded in absolute value so that:

${\lim}_{n \to \infty} {s}_{n}$

does not exist.