Variable substitution & Universal Quantifier: How to arrive from $\forall n > N.|a_n-L|< \epsilon$ to $\forall n > N.| a_{n-1}-L | < \epsilon$?

Question

I have trouble understanding variable substitution in universal statement regarding the decrementation of an index:

We have the definition: $ \forall \epsilon>0 \, \exists N\in \mathbb{N} \, \forall n > N . | a_n - L | < \epsilon. $
I am given $ \epsilon > 0 $ , so it means there exists $ N \in \mathbb{N} $ s.t. $ \forall n > N . | a_n - L | < \epsilon. $

Question: I know it now follows that $ \forall n > N . | a_{n+1} - L | < \epsilon. $ because of the following reasoning: Let $ n>N $ be arbitrary, $ n+1>N $ so ( by universal instantiation ) $ | a_{n+1} - L | < \epsilon. $ Since $ n $ was arbitrary, therefore $ \forall n > N . | a_{n+1} - L | < \epsilon. $
But I've also encountered places where it is written that $ \forall n > N . | a_{n-1} - L | < \epsilon $ and I don't know how to arrive at this universal statement using logical steps like I showed when I arrived to $ \forall n > N . | a_{n+1} - L | < \epsilon. $
Can you please show how to arrive to $ \forall n > N . | a_{n-1} - L | < \epsilon $ ?

For example, when learning about sums In class we had the following: ( $ S_n $ is the partial sum of sequence $ (a_n) $, we're given $ \epsilon > 0$ and that $ N \in \mathbb{N} $ exists) $ \forall n > N . | S_n - L | < \epsilon/2. $ Therefore for all $n>N $ we have$ |S_n - S_{n-1} | = | S_n - L + L - S_{n-1} | \leq | S_n - L | + | S_{n-1} - L | = \epsilon/2 + \epsilon/2 = \epsilon $.
And I just didn't know how to show $ \forall n > N . | S_{n-1} - L | < \epsilon/2. $
( Although I could show that $ \forall n > N . | S_{n+1} - L | < \epsilon/2 $ and arrive to the same result just by deriving the steps from $ |S_{n+1} - S_{n} |, $ I still want to know how to arrive to $ \forall n > N . | S_{n-1} - L | < \epsilon/2. )$

Answer 1

$|S_n-L|<\epsilon/2$ for $n>N$ does $\textbf{not}$ imply that $|S_{n-1}-L|<\epsilon/2$ for $n>N$.

However, if you can pick $N$ such that $|S_n-L|<\epsilon/2$ for $n>N$, then you can also pick $M$ such that $|S_{n-1}-L|<\epsilon/2$ for $n>M$. In fact, you can just take $M=N+1$. Sometimes you abuse notation and just call everything $N$ but it seems to me that this is what's being done.