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

  • $\begingroup$ 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. $\endgroup$ – Henno Brandsma Mar 26 at 9:19
  • $\begingroup$ @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 $\endgroup$ – 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.

  • $\begingroup$ Can you please explain clearly the " second implies first " section $\endgroup$ – PN Das Mar 26 at 9:39
  • $\begingroup$ But thank you for your effort I will try to figure this out $\endgroup$ – PN Das Mar 26 at 9:39
  • $\begingroup$ 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. $\endgroup$ – 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.


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