# Changing a finite number of terms in a sequence does not effect the limit

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 $-\infty.$ $\blacksquare$

• Why do you keep changing the tags? – Bungo Apr 6 '16 at 2:05
• ....yes but. What if the finite terms we change are bigger than N? – fleablood Apr 6 '16 at 2:40
• @fleablood N is a variable that represents an $a_N$ beyond which there are no finite terms that change as I understand it. If there are we can then choose $N$ accordingly so that all different finite terms are in the sequence as less than $a_n$. – Red Apr 6 '16 at 23:39
• Precise definitions for proofs can be frustrating. The "idea" is that if there are only a finite changes of difference then there are infinite that are the same and the finite ones "don't really matter". Because it's the "tail" that determines the limit And the tail can "always" "start" beyond any specific finite number of terms. But try putting that into correct terms... – fleablood Apr 7 '16 at 18:18
• I definitely knew what you were going for but I don't think you specifically addressed it. The n < N do not affect the limit. True. But you need to point out that there are only a finite number n that are changed (we should probably use the variable i rather than n). The i < N do not affect the limit but we need to point out that as there are only a finite number of i that are changed, that we can find an N > any of the i. That's all. That seemed to be lacking in your proof. Or so is my opinion. – fleablood Apr 7 '16 at 18:23

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 $A$.
• Can you necessarily say that the sequences must agree, i.e. $a_n=b_n$, when $n >M$, for some $M$? What if the very last terms are different? Or maybe there's no "last term", since the sequences are infinite? – mavavilj Oct 1 '16 at 17:24
• @mavavilj There is no last term. If you tell me you found the „last“ term, say $a_n$, I can always find a later one $a_{n+1}$, or I should rather say infinitely many later ones. That is the whole weirdness of the concept of infinity. – philmcole Jan 7 '18 at 19:03