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?
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.