Evaluate the summation of $(-1)^k$ from $k=0$ to $k=n$

Question

$$\sum_{k=0}^n(-1)^k$$

I know that the answer will be either -1 or 0 depending on whether there are an odd or an even number of sums in total, but how can I determine this if $k$ goes to infinity (which I am thinking means there is neither an even nor odd amount of sums).

How would I determine this?

The series does not converge: when $n$ goes to infinity, the sum does not "go" to anything, at least not in the usual sense of limits. — Clement C., Nov 07 '16 at 18:38
The partial sums oscillate between $-1$ and $0$ , as you said. so the limit does not exist. — Peter, Nov 07 '16 at 18:38
The series $\sum_{k\ge 0}(-1)^k$ is divergent: it has no defined value. — Brian M. Scott, Nov 07 '16 at 18:39
Since you're starting with $k=0$ and $(-1)^0=1$, the sum will oscillate between $1$ and $0$, not $-1$ and $0$. — Barry Cipra, Nov 07 '16 at 18:41

score 3 · Accepted Answer · answered Nov 07 '16 at 19:48

Using the formula for the sum of a geometric series, we get $$ \begin{align} \sum_{k=0}^n(-1)^n &=\frac{1-(-1)^{n+1}}{1-(-1)}\\ &=\frac{1+(-1)^n}2\tag{1} \end{align} $$ Thus, the sum does not converge to a limit. However, if we consider the Cesàro Sum of this series, we get $$ \begin{align} \lim_{n\to\infty}\frac1n\sum_{k=1}^ns_k &=\lim_{n\to\infty}\frac1n\sum_{k=1}^n\frac{1+(-1)^k}2\\ &=\lim_{n\to\infty}\frac1n\left(\frac n2-\frac{1-(-1)^n}4\right)\\[3pt] &=\frac12\tag{2} \end{align} $$

score 1 · Answer 2 · answered Nov 07 '16 at 19:29

1

Using various summation methods you can arrive at $1/2$ being the regularized sum at infinity.

answered Nov 07 '16 at 19:29

Anixx

9,119

score 0 · Answer 3 · answered Nov 07 '16 at 18:54

0

$\sum_{k=0}^n (-1)^k = {1 \over 2} (1+(-1)^{n})$.

answered Nov 07 '16 at 18:54

copper.hat

172,524

Why the downvote? – copper.hat Dec 10 '16 at 17:33

score 0 · Answer 4 · answered Nov 07 '16 at 19:26

Using the general form derived as: \begin{align} \sum_{k=0}^{n} x^{k} &= \sum_{k=0}^{\infty} x^{k} - \sum_{k=n+1}^{\infty} x^{k} = \frac{1}{1-x} - \frac{x^{n+1}}{1-x} = \frac{1 - x^{n+1}}{1-x} \end{align} then for $x = -1$ the result becomes $$ \sum_{k=0}^{n} (-1)^{k} = \frac{1 + (-1)^{n}}{2}.$$

Evaluate the summation of $(-1)^k$ from $k=0$ to $k=n$

4 Answers4