# Question on Cauchy Criterion of Series

We know that the Cauchy Criterion of a series is as follow (proof is taken as excerpt from an analysis book):

Theorem: A series $$\sum_{j=1}^{\infty}a_j$$ converges iff for all $$\epsilon>0$$ there is an $$N\in \mathbb{N}$$ so that for all $$n\ge m \ge N$$ we have $$|\sum_{j=m}^{n} a_j|< \epsilon$$.

Let $$s_n$$ denotes partial sum $$\sum_{j=1}^n a_j$$.

We know that the proof of $$\Rightarrow$$ direction makes use of the fact that convergent series implies the sequence of partial sum converges and thus is Cauchy in $$\mathbb{R}$$. So here is part of the proof, since sequence of partial sum is Cauchy, therefore given $$\epsilon>0$$, there exists $$N \in \mathbb{N}$$ such that for all $$n \ge m \ge N$$, $$|s_n - s_{m-1}|< \epsilon$$.

I do not get the $$s_{m-1}$$ part, i.e. I do not get why you can pick $$m-1$$, since there is possibility for $$m=N$$, and thus having $$m-1, which in this case you cannot guarantee the sequence of partial sum Cauchy.

1. Starting from the sequence of partial sum being Cauchy, $$|s_n-s_m|< \epsilon$$, for all $$n,m\ge N$$ (the definition of Cauchy sequence), and without loss of generality we can pick $$n>m$$ such that $$|s_n- s_m|=|\sum_{j=m+1}^{n} a_j|< \epsilon$$, but this differs from the statement of the theorem somehow.
2. I was also thinking of having $$m \ne N$$ and $$m>N$$ so that we have $$m-1\ge N$$ and that the original proof of the theorem that I showed right after the theorem is valid.
• Pick your $N$ such that $n, m \geq N$ implies $|s_n - s_m| < \varepsilon$. Then use $N + 1$ instead. I agree that this is not symbol for symbol how one defines Cauchy, but playing little games with indices is something one has to get used to. – Dylan Moreland Jul 18 '12 at 21:52
It makes no difference at all if you use $n\geq m > N$ in the definition of Cauchy sequnces, so I guess your 2 option is a good solution to your problems. It is just a problem of what name you give to the numbers. If the sequence of partial sums is a Cauchy sequences there exists such $N$ as you defined. Now call $N' = N+1$ and conclude what you want.
• Shouldn't it be there exists such $N'$ – Daniel Jul 18 '12 at 22:07