Prove the alternating series test from the Cauchy convergence theorem for series. We want to show $\sum_{k=1}^{\infty} (-1)^{k+1}a_k$ converges.
Proof:
Let $\epsilon>0$. Let $k,k'\in\mathbb{N}$ and let $a_k$ be a decreasing monotone sequence of non negative numbers such that $a_k\rightarrow 0$. It follows by definition of convergence that there exists a $N'\in\mathbb{N}$ such that if $k\geq N'$ then $|a_k|=a_k<\frac{\epsilon}{k'}$. Now let $N=N'$ such that if $n>N'$ then $|(-1)^{n+2}a_{n+1}+….+(-1)^{n+k'+1}a_{n+k'}|\leq |(-1)^{n+2}a_{n+1}|+|(-1)^{n+3}a_{n+2}|+…+|(-1)^{n+k'+1}a_{n+k'}|=a_{n+1}+..+a_{n+k'}\leq (k')(a_{n+1})<k'*\frac{\epsilon}{k'}=\epsilon$. Thus the series converges. Is this correct? The only issue I could see is the $k'$ but since its fixed I'm not sure. I just want to know if my proof is correct or what can i do to correct it?
Best Answer
Some intuition why your proof isn't quite correct: When you used the triangle inequality $$|(-1)^{n+2}a_{n+1} + \cdots + (-1)^{n+k'+1}a_{n+k'}| \leq |(-1)^{n+2}a_{n+1}| + \cdots + |(-1)^{n+k'+1}a_{n+k'}|\\ = a_{n+1} + \cdots + a_{n+k'} $$ you removed the alternating nature of the series, so if your proof was correct, you would have proved convergence for the non-alternating series $\sum a_k$ as well.
The technical trouble comes from the fact that to show Cauchy, you need to be able to select any $n, m \geq N$ ($m$ and $n$ can not depend on each other, for example), but by selecting $k'$ beforehand and then letting $N$ depend on $k'$, you force $m = n+k'$ to depend on $n$ through your selection of $N$. (Does this make sense?).
Here's an argument to help you prove what you want. Let $b_k = (-1)^k$ and let's note that for any $m \leq n$ $$\Big| \sum_{k=m}^n b_k \Big| \leq 1$$ From a telescoping series argument, you can show that $a_k = \sum_{l=m}^{k-1} (a_{l+1} - a_{l}) + a_m$. Therefore, $$ \sum_{k=m}^n b_k a_k = \sum_{k=m}^n b_k \Big( \sum_{l=m}^{k-1} (a_{l+1} - a_l) + a_m \Big) = \sum_{l=m}^{n-1} \Big\{ (a_{l+1} - a_l) \sum_{k=l}^n b_k \Big\} + a_m \sum_{k=m}^n b_k $$ You can now deduce that $$ \Big| \sum_{k=m}^n b_k a_k \Big| \leq \sum_{l=m}^{n-1} \Big\{ |a_{l+1} - a_l| \Big|\sum_{k=l}^n b_k \Big| \Big\} + a_m \Big| \sum_{k=m}^n b_k \Big| \leq \sum_{l=m}^{n-1}|a_{l+1} - a_l| + a_m. $$ Since $a_l \geq a_{l+1}$ for every $k$, $|a_{l+1} - a_l| = a_l - a_{l+1}$, which reduces our last inequality to $$ \Big| \sum_{k=m}^n b_k a_k \Big| \leq \sum_{l=m}^{n-1} (a_{l} - a_{l+1}) + a_m = 2a_m - a_n. $$ where the last equality used another telescoping series argument to show $\sum_{l=m}^{n-1} (a_{l} - a_{l+1}) = a_m - a_{n}$. Now, had you originally chosen $N \in \Bbb{N}$ with $n, m \geq N$, then (by monotonicity) $a_m \leq a_N$ and also $a_n \geq 0$, so you finally arrive at $$ \Big| \sum_{k=m}^n b_k a_k \Big| \leq 2a_m - a_n \leq 2 a_N - 0 = 2a_N. $$ So, at last, if $\epsilon > 0$ and you choose $N$ such that $a_N < \epsilon /2$, what all this work has shown is that if $N \leq m, n$ (and w.l.o.g. $m < n$) then
$$ \Big| \sum_{k=m}^n b_k a_k \Big| \leq 2a_N < 2(\epsilon/2) = \epsilon. $$