Show If $a_n\sim b_n$ and $c_n \sim d_n$ then $a_n+c_n\sim b_n+d_n$. $a_n\sim b_n$ means $a_n/b_n \rightarrow 1$.
I set up the definition, but the addition in the denominator causes problems.
$\big|a_n/(b_n+d_n)+c_n/(b_n+d_n)-1\big|<\epsilon$
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This isn't true, consider $a_n = 1$, $b_n = 1$, $d_n = -\frac{1}{n} - 1$ and $c_n = \frac{1}{n} - 1$.
Clearly $a_n \sim b_n$. On the other hand
$$\frac{c_n}{d_n} = \frac{1/n-1}{-1/n-1} \to \frac{-1}{-1} = 1.$$
Thus $c_n \sim d_n$.
On the other hand $a_n + c_n = 1/n$ and $b_n + d_n = -1/n$. Thus,
$$\frac{a_n + c_n}{b_n + d_n} = \frac{1/n}{-1/n} = -1 \not\to 1.$$
If you assume that $a_n, b_n, c_n, d_n > 0$. Then $a_n \sim b_n$ means for all $\epsilon > 0$ for sufficiently large $n$ we have that $|a_n - b_n| < \epsilon|b_n|$. Using this we get that
$$\left|\frac{a_n+c_n}{b_n+d_n} - 1\right| = \frac{|a_n-b_n+c_n-d_n|}{|b_n+d_n|} \le \frac{|a_n - b_n|+|c_n - d_n|}{|b_n + d_n|} < \epsilon \frac{|b_n|+|d_n|}{|b_n + d_n|} = \epsilon.$$
Where the last equality comes for the positivity assumption.
Rishi Sonthalia
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