# How these two definitions are equivalent?

Please have a look at these two equivalent defintions of Cauchy's general principle of convergence series. I understand the first defintion but I'm having problems with the second defintion.

First definition says a series $$\sum a_n$$ is convergent iff for each $$\epsilon \gt 0$$ $$\exists$$ a number $$N \in \mathbb N$$ such that $$m , n \gt N$$ implies $$|S_n - S_m| \lt \epsilon$$.

Since $$S_n = \sum_{k=1}^n {a_k}$$ and $$S_m = \sum_{k=1}^m {a_k}$$ $$|S_n - S_m| = \sum_{m+1}^{n} {a_k}$$ ( taking $$n\ge m$$ ).

So the first statement can be restated as ( I think ) a series $$\sum a_n$$ is convergent iff for each $$\epsilon \gt 0$$ $$\exists$$ a number $$N \in \mathbb N$$ such that $$n \ge m \gt N$$ implies $$|\sum_{m+1}^{n} {a_k}| \lt \epsilon$$.

Second definition says a series $$\sum a_n$$ is convergent iff for each $$\epsilon \gt 0$$ $$\exists$$ a number $$N \in \mathbb N$$ such that $$n \ge m \gt N$$ implies $$|S_n- S_{m-1}| \lt \epsilon$$.

Similiarly this can be restated as a series $$\sum a_n$$ is convergent iff for each $$\epsilon \gt 0$$ $$\exists$$ a number $$N \in \mathbb N$$ such that $$n \ge m \gt N$$ implies $$|\sum_{m}^{n} {a_k}| \lt \epsilon$$.

How these definitions are equivalent ? I want to know if the second definition is really true. How do I prove the second one?

First definition

Second definition

• The texts define the "Cauchy criterion". You cannot prove a definition, but you can show that if a series obeys definition 1. it also obeys definition 2. and vice versa and in that sense you can say that definitions can be equivalent. The theorem then is that a series obeying that criterion is convergent. The statement of convergence is not part of the definition, as you suggest. – Henno Brandsma Mar 26 at 9:19
• @HennoBrandsma we know a series is convergent iff $| \sum_{m+1}^{n} {a_k}| \lt \epsilon$. Which can be obtained from first definition. But if we simplify second defintion then it would be a series is convergent iff $| \sum_m^n {a_k} | \lt \epsilon$. But these two things seems to be different but as per many books thaey are same – PN Das Mar 26 at 9:29

They are the same because the $$N$$ , which needs to exist for each $$\epsilon$$, can be chosen arbitrarily large due to the given conditions.

That is, if some $$N_0$$ "works as per first/second definition" for a given $$\epsilon$$ (i.e. $$N_0$$ satisfies : if $$m,n > N_0$$ then $$|S_m - S_n| < \epsilon$$), then even $$N_0+1,N_0+2,...$$ all "work" for the $$\epsilon$$.

Now, you can use this flexibility to get over the restrictions imposed by the second definition.

First implies second : Let $$\sum a_n$$ converge as per the first definition. We want to show that it converges as per the second definition. Start with $$\epsilon > 0$$.

Then, there is $$N >0$$ as per first definition. Let $$N_0 = N+1$$. Then, if $$n \geq m > N_0$$, we see that $$n > N$$ and $$m-1 > N$$, so $$|S_n - S_{m-1}| < \epsilon$$ by the first definition. Since for every $$\epsilon$$ we have found $$N_0$$, $$\sum a_n$$ converges as per second definition.

Second implies first : Again, fix $$\epsilon > 0$$. Get an $$N$$ as per second definition and again define $$N_0 = N+1$$. Now, for any $$n ,m$$, we have $$|S_n - S_{m-1}| = |S_{m-1} - S_n|$$, since the arguments are just additive inverses of each other.

Now if $$n,m > N_0$$ then $$n > N$$, and $$m-1 > N$$. Now, one of three things happens :

• If $$n \geq m$$ then $$|S_{n} - S_{m-1}| < \epsilon$$.

• If $$n = m-1$$ then $$|S_n - S_{m-1}| = 0 < \epsilon$$.

• If $$n < m-1$$ then $$m-1 \geq n+1$$ so $$|S_{m-1} - S_{(n+1)-1}| < \epsilon$$.

Either way, the given $$N_0$$ works and we are done.

Note : We have used the fact that if $$N$$ works then $$N+1$$ works. While taking maxima of various $$N$$s, as you may do later in your course, you will realize that the property that you can take $$N$$ finite but arbitrarily large is helpful for proof writing.

• Can you please explain clearly the " second implies first " section – PN Das Mar 26 at 9:39
• But thank you for your effort I will try to figure this out – PN Das Mar 26 at 9:39
• The thing is, that the second condition is almost symmetric because $|S_k - S_l| = |S_l - S_k|$ for any $l,k$. I am exploiting that , along with just adjusting the indices $n$ and $m$ slightly so as to satisfy the second conditions. – астон вілла олоф мэллбэрг Mar 26 at 9:44

Yes, they are equivalent. Suppose that the first condition holds. If $$\varepsilon>0$$, there is a natural $$N$$ such that$$m,n>N\implies\lvert S_n-S_m\rvert<\varepsilon.$$So, if $$n\geqslant m>N+1$$, then both numbers $$n$$ and $$m-1$$ are greater than $$N$$ and therefore $$\lvert S_n-S_{m-1}\rvert<\varepsilon$$. The other direction is similar.