# Proving that if the sequence $\{s_n-L\}$ converges to zero, then a sequence $\{s_n\}$ converges to a limit $L$

I am having trouble proving this statement without using the limit rules. I know I start by assuming that the sequence $\{s_n-L\}$ converges to zero, therefore, for every number $ϵ > 0$, there is an integer $N$ such that $\|s_n-L-0\|$ $< ϵ$, whenever n > N. How would I prove that $\lim_{n\to\infty}(s_n)=L$ without using the limit subtraction rule and just the definition of a limit?

HINT:

You stated the definition of the limit:

For all $\epsilon >0$ there is a number $N$ so that

$$-\epsilon <s_n-L<\epsilon \tag 1$$

whenever $n>N$.

Now, add $L$ from both sides of $(1)$. What can you conclude?

• We conclude that $−ϵ+L < s_n < ϵ+L$, therefore $s_n$ converges to $L$? – Larry Feb 13 '16 at 2:07
• Yes, that is correct. Well done. – Mark Viola Feb 13 '16 at 2:08

It is a direct consequence of the definition of limit which I repeat for clarity:

Definition: A sequence $\{a_{n}\}$ tends to a limit $A$ as $n \to \infty$ if for any arbitrary number $\epsilon > 0$ there is a positive integer $N$ such that $$|a_{n} - A| < \epsilon$$ whenever $n \geq N$.

We are given that $\{s_{n} - L\}$ tends to $0$ as $n \to \infty$ and hence by the above definition it follows that for any arbitrary number $\epsilon > 0$ there is a positive integer $N$ such that $$|(s_{n} - L) - 0| < \epsilon$$ whenever $n \geq N$.

The above statement is same as the following

For any arbitrary number $\epsilon > 0$ there is a positive integer $N$ such that $$|s_{n} - L| < \epsilon$$ whenever $n \geq N$.

and comparing this with the definition given in the beginning we see that $\lim_{n \to \infty}s_{n} = L$.