I have the simple geometric series $\Sigma\,x_n$ where $x_n = (-1)^n$ so that $\Sigma\,x_n$ diverges. However, I know that the sequence $(\sigma_n)_n = \frac{1}{n}\Sigma\,s_n$ converges, where $s_n$ are the partial sums $x_1+x_2+…+x_n$, and that $lim\,\sigma_n$ exists.
I can almost prove this and find the limit as follows:
$s_n = \Sigma\,x_n = -1 + 1 – 1 + 1 – 1\,\,+\,\,…\,+\,(-1)^n$
And so
$\sigma_n = \frac{1}{n}(-1+0-1+0-1\,\,+\,\,…\,+\,\Sigma\,x_n)$
Which results in a series of the form:
$(\sigma_n)_n= (-1, -\frac{1}{2}, -\frac{2}{3}, -\frac{1}{2}, -\frac{3}{5}, -\frac{1}{2}, \,\,…)$
Such that I can infer the following pattern (apologies for my present inability to properly represent piecewise functions):
$\sigma_n= \{^{-\frac{1}{2},\,\,n\,even}_{-(\frac{1}{2} + \frac{1}{2n}),\,\,n\,odd} $
From here, it's obvious that the sequence has a total of two subsequential limits that are both equal to $-\frac{1}{2}$, so $lim\,\sigma_n=-\frac{1}{2}$.
However, I can't figure out a satisfying way to properly prove the sequence is equal to what I've inferred. Is there something I can do beyond just calculating enough terms to make a good guess? What am I missing?
Best Answer
$s_n =\sum_{k=1}^n (-1)^k$ so $s_{2n} =\sum_{k=1}^{2n} (-1)^k =\sum_{k=1}^{n} ((-1)^{2k-1}+(-1)^{2k} =0 $ and $s_{2n+1} =s_{2n}+(-1)^{2n+1} =-1 $.
Therefore, if $\sigma_n =\frac1{n}\sum_{k=1}^n s_k $ then
$\begin{array}\\ \sigma_{2n} &=\frac1{2n}\sum_{k=1}^{2n} s_k\\ &=\frac1{2n}\sum_{k=1}^{n} (s_{2k-1}+s_{2k})\\ &=\frac1{2n}\sum_{k=1}^{n} (-1+0)\\ &=-\frac12\\ \end{array}$
We also have
$\begin{array}\\ \sigma_{2n+1} &=\frac1{2n+1}\sum_{k=1}^{2n+1} s_k\\ &=\frac1{2n+1}(\sum_{k=1}^{2n} s_k+s_{2n+1})\\ &=\frac1{2n+1}(2n\sigma_{2n}+s_{2n+1})\\ &=\frac1{2n+1}(-2n\frac12-1)\\ &=\frac1{2n+1}(-n-1)\\ &=-\frac{n+1}{2n+1}\\ &=-\frac{n+1/2+1/2}{2n+1}\\ &=-\frac12-\frac{1}{4n+2}\\ \end{array} $
This is more than enough to prove that $\lim_{n \to \infty} \sigma_n =-\frac12 $.