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!
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. $0, 1, 0, 1, \dots$.