[Math] Changing a finite number of terms in a sequence does not effect the limit

analysislimitsproof-verificationreal-analysissequences-and-series

In Introduction to Infinite Series by Bonar and Khoury, the following are given as "facts" but left up to the reader to prove. Can you please let me know if I made any errors in my attempts and if the proof is completely wrong give the correct proof? Also you are more than welcomed to give a shorter proof.

Theorem:

Changing a finite number of terms in a sequence has no effect on the convergence, divergence or the limit if it exists.

For example, the sequences

$$1, \frac{1}{2},\frac{1}{3},\frac{1}{4},\frac{1}{5},\frac{1}{6},\frac{1}{7}\cdots , \frac{1}{n}, \cdots$$

and

$$ 2,7,5,\frac{1}{10},\frac{1}{5},\frac{1}{6}, \frac{1}{7}, \cdots , \frac{1}{n}, \cdots$$

both converge $0$.

Proof. By definition, a sequence converges to a real number $A$ if , for each $\epsilon>0$, there exists an integer $N$ such that for all $n>N$, $|a_n-A|<\epsilon$. As we can see from the definition changing any terms $|a_N|$ for all $N<n$ does not affect the limit and therefore convergence because $|a_N|$ is independent on the choice of $\epsilon$.

And by definition, a sequence diverges to $\infty$ if, for any $M>0$, there exists an integer $N$ such that all $n>N$ , $a_n>M$. Similarly, we can change any $a_N<M$, and it does not affect the choice of $\epsilon$ for $a_n$.A similar argument can be used for the case of divergence to $-\infty.$ $\blacksquare$

Best Answer

Your idea is right. You can make it a bit more formal as follows.

Suppose we start with a sequence $a_n$. If we change finitely many terms, then this results in a new sequence $b_n$. Since we only changed finitely many terms, there is some $M$ such that $a_n = b_n$ for all $n > M$.

Now suppose that $a_n$ converges to $A$. Let $\epsilon > 0$. There is some $N$ such that $|a_n - A| < \epsilon$ for all $n > N$.

Then, for all $n > \max\{N, M\}$, we have $b_n = a_n$, so $$|b_n - A| = |a_n - A| < \epsilon$$ which shows that $b_n$ also converges to $A$.

We have shown that if two sequences differ in only finitely many terms, and one sequence converges, then the other also converges (to the same limit). The contrapositive: if one diverges, then the other must also diverge.

Related Question