Take the series $\sum_{k=1}^\infty \frac{(-1)^k}{2^k+k}$
Since I use the Cauchy Criterion, for $n, m$ ∈ ℕ with $m > n$, define:
$S_{m,n} = \sum_{k=n+1}^m \frac{(-1)^k}{2^k+k}$
Here is my (shortened) proof:
If $n$ and $m$ are odd then
- $S_{m,n} = \frac{1}{2^{n+1}+n+1} - \frac{1}{2^{n+2}+n+2}$) + ... + ($\frac{1}{2^{m-1}+m-1} - \frac{1}{2^{m}+m}$)
All terms are positive because the series is decreasing and in each bracket, the next term is being taken away from the previous one, so $S_{m,n} \geq 0$ (This explanation feels horrible, please correct me on how to properly word this)
- $S$$n,m$ = $\frac{1}{2^{n+1}+n+1}$ + ($-\frac{1}{2^{n+2}+n+2} + \frac{1}{2^{n+3}+n+3}$) + ... + ($-\frac{1}{2^{m-2}+m-2} + \frac{1}{2^{m-1}+m-1}$) - $\frac{1}{2^{m}+m}$
All brackets are negative in addition to the final term of the series being negative, so $S_{m,n} \leq\frac{1}{2^{n+1}+n+1}$
Note, this is just the case when $n$ and $m$ are odd and I know I'd have to do the rest of the cases. I just wanted to make sure my general logic was sound
So $0 \leq S_{m,n} \leq\frac{1}{2^{n+1}+n+1}$
At the end of all cases, you would get:
- $\vert S_{m,n}\vert \leq \frac{1}{2^{n+1}+n+1}$
Now time to finish it off and apply the Cauchy Criterion. Given $\varepsilon > 0$, let $N = \lceil \frac{1}{\varepsilon}\rceil$.
Then we have for all $m > n > N$:
- $\vert S_{m,n}\vert\leq\frac{1}{2^{n+1}+n+1}$ < $\frac{1}{N}\leq\varepsilon$
So the series exists and it converges. Proof done? This is my first module on real analysis so please understand that I'm having difficulty with the general topic. Thank you.