Proving convergence using the difference of series. We are given the convergent series $$\sum_{n=1}^{+\infty}a_n$$
and the additional series $$\sum_{n=1}^{+\infty}b_n$$ such that $$\lim_{n \to \infty} |a_n - b_n| = 0.$$ I am looking to prove that $$\sum_{n=1}^{+\infty}b_n$$ converges.
I have been told that the sum of convergent series is convergent, but is there a similar theorem for the difference of convergent series? If it's as simple as "this series $b_n$ cannot converge," what is an example of sequences that disprove the idea?
 A: As the user amsmath pointed out in their comment, the condition $\lim_{n\to\infty}|a_n-b_n|$ does not control the rate of decay of the sequence $b_n$. Since $\sum a_n$ converges, $a_n\to 0$ in particular, so no matter which sequence $b_n$ is, if $b_n\to 0$, then $\lim_{n\to\infty}|a_n-b_n| = 0$. For example, $b_n = n^{-1}$ is possible, and of course $\sum n^{-1}$ diverges.
Suppose for simplicity that $a_n, b_n > 0$. In this case, if you work with an analogous condition, but with quotients instead of differences, namely that $$\lim_{n\to\infty}\frac{b_n}{a_n} = 1,\tag{1}$$ then this condition does control the rate of decay of $b_n$, and in particular, if $\sum a_n$ converges, so does $\sum b_n$ because,
$$
\sum b_n = \sum a_n\cdot\frac{b_n}{a_n} \le C\sum a_n,
$$
for an appropriate choice of the constant $C$.
The example of $b_n = n^{-1}$ suggests a possible modification to the difference condition, namely, what if we suppose instead that
$$
\lim_{n\to\infty}n|a_n-b_n| = 0,
$$
i.e., we ask the same question as to whether $\sum b_n$ must converge now. This new condition tries to say that $b_n$ decays faster than $n^{-1}$.  It turns out the answer is still no, because of a related series $\sum b_n:=\sum (n\log n)^{-1}$. However, the condition
$$
\lim_{n\to\infty}n^{1+\epsilon}|a_n-b_n| = 0,
$$
for any positive $\epsilon > 0$, would be enough to guarantee that $\sum b_n$ converges, assuming $\sum a_n$ converges. However, the utility of a result like this is limited in scope. To illustrate, again assume $a_n,b_n>0$. What a condition like $\lim_{n\to\infty}n^{1+\epsilon}|a_n-b_n|= 0$ is trying to argue, morally, is that because $a_n$ is convergent, and because $b_n$ is "close" to $a_n$, $b_n$ is also convergent. However, since the series $a_n$ could go to zero at a very fast rate, any "weight" like $n^{1+\epsilon}$ is only going to tell us something meaningful about $b_n$ with regards to $a_n$ if $a_n$ decays no faster than $1/n^{1+\epsilon}$.
To capture meaningful information about $b_n$ with regards to $a_n$ for an arbitrary $a_n$ such that $\sum a_n$ converges, we could take as our weight $\frac{1}{a_n}$ itself(!), since then we can say that $b_n$ is getting arbitrarily close to $a_n$, in a way that is genuinely meaningful with regards to the sequence $a_n$, regardless of what $a_n$ is. Written out, this becomes the condition
$$
0 = \lim_{n\to\infty}\frac{1}{a_n}|a_n-b_n| = \lim_{n\to\infty}|1-\frac{b_n}{a_n}|,
$$
which is equivalent to the condition $(1)$. This is a form of the limit comparison test for convergence.
