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Let $(x_n)$ be a Cauchy sequence in R and $(y_n)$ is a sequence in R such that $|x_n-y_n|<\frac{1}{n}$ for all $n\geq 1$. Prove that $y_n$ is a Cauchy sequence and $lim (x_n)=lim (y_n)$

My attempt (rough sketch):

Assume $y_n$ is a sequence in R. If $y_n$ is a convergent sequence then $y_n$ is a Cauchy sequence.

Suppose $y_n$ is divergent.

Then either it diverges to positive infinity or to negative infinity.

If it diverges to positive infinity then for a pre- assigned positive number G,however large, there exists a natural number k such that $y_n> G$ for all $n\geq k$.

But given that $|x_n-y_n|<\frac{1}{n}$

This implies, $ \frac {-1}{n} +x_n<y_n<\frac{1}{n}+x_n$. But $x_n$ is bounded and so is $1/n$. This implies that $y_n$ is bounded. Which is contradiction to assumption that $y_n$ is divergent. Hence, $y_n$ is convergent. Which implies that it is a Cauchy sequence.

Is this correct? Thanks in advance!

Natasha J
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1 Answers1

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There is no need to look at separate cases or use proof by contradiction. The inequality you give

$$\frac {-1}{n} +x_n<y_n<\frac{1}{n}+x_n$$

is enough to prove the result via the squeeze theorem (using the fact that convergent sequences and Cauchy sequences are the same thing in $\mathbb R$).

By the way, if you do look at separate cases, $y_n$ could also diverge by being oscillatory, e.g. $0, 1, 0, 1, \dots$.

Linear Christmas
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  • Thanks for answering. So other than being oscillating sequence, is the proof correct? – Natasha J Apr 24 '21 at 10:10
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    @NatashaJ No problem. If you somehow also rule out oscillating sequences (notice that boundedness will not give a contradiction here), you still need to prove that the limit of $(y_n)$ is the same as the limit of $(x_n)$. The most natural way to do this, especially since you already noticed the inequalities, is to use the squeeze theorem. Once you prove the final part with the squeeze theorem, you may note that the entire theorem boils down to the inequalites and an application of the squeeze theorem; so one may do a direct proof instead of proof by cases + contradiction. – Linear Christmas Apr 24 '21 at 10:17
  • Thank you so much! I got it now!! – Natasha J Apr 24 '21 at 10:19