# Find the sequence of partial sums for the series $a_n = (-1)^n$ Does this series converge?

Find the sequence of partial sums for the series $$\sum_{n=0}^\infty (-1)^n = 1 -1 + 1 -1 + 1 - \cdots$$

Does this series converge ?

My answer is that the sequence $= 0.5 + 0.5(-1)^n$. This makes a sequence that alternates between $1$ and $0$.

I know that the sequence does not converge since it is not monotone. But how can I prove this?

• Do there exist subsequences which converge to disparate limits? – abiessu May 29 '15 at 0:00
• – Victor May 29 '15 at 0:02
• A series does not have to be monotone to converge, e.g., $(-1)^n/n$. What theorems/tests do you have for convergence? Cauchy? For instance, if $S_n$ is the partial sum, then $|S_n - S_{n+1}| = 1$ and hence is not Cauchy. – Simon S May 29 '15 at 0:02

hint: $$s_{2n} = 0, s_{2n+1} = 1$$.
If it converges to $\ell$ then $|s_n-\ell|<\varepsilon$, once $\varepsilon>0$ is fixed, and $n$ is sufficiently large. Here $s_n$ is exactly the $n$-th partial sum.
But if you set, for example, $\ell=1/10$, then the above inequality cannot be satisfied even for $n$ enough large.
All you need to show is that the the sequence $x_{n}=(-1)^{n}$ does not converge. Many ways to do it : First, using the theorem saying that if a sequence converges , then any subsequence converges to the same limit. $x_{2n}$ and $x_{2n+1}$ are subsequences converging obviously to different limits. Not familiar with this? No problem! Assume that $x_{n}$ converges, say to $l$. Then, given $\epsilon >0$, we can find an integer $N(\epsilon)$ such that $n \geq N(\epsilon)$ implies $|x_{n}-l| <\epsilon.$ Now, $|x_{n+1}-x_{n}| = |x_{n+1}-l+l-x_{n}| \leq |x_{n+1}-l|+|l-x_{n}| <2\epsilon$, provided $n \geq N(\epsilon)$. Now, we chose $\epsilon=1$. And I will leave it to you to see the obvious contradiction!.
$$\sum\limits_{n=0}^\infty (-1)^n = \lim\limits_{m\to\infty}\sum\limits_{n=0}^m (-1)^n$$ $$= \lim\limits_{m\to\infty} \frac12\left((-1)^m + 1\right)$$ Note that the sequence $$a_m=\frac12\left((-1)^m + 1\right)$$ is convergent if $$\lim\limits_{m\to\infty} a_{2m} = \lim\limits_{m\to\infty} a_{2m+1} = L$$ However, in this case we have $$\lim\limits_{m\to\infty} \frac12\left((-1)^{2m}+1\right) =\frac22= 1$$ And $$\lim\limits_{m\to\infty} \frac12\left((-1)^{2m+1}+1\right) =\frac02= 0$$ Therefore $a_m$ is a divergent sequence.