I am trying to solve question ($6$) in section $6.11$ of Goldblatt's Lectures on the Hyperreals.
The question asks:
Given two series of positive terms $\sum_1^\infty a_i$ and $\sum_1^\infty b_i$ such that the sequence $(a_i/b_i:i\in\mathbb{N})$ converges in $\mathbb{R}$, show that for unlimited $m$ and $n$, $\sum_m^n a_i$ is infinitesimal if and only if $\sum_m^n b_i$ is infitesimal.
I have not got very far.
I tried to first show the forward implication. For all naturals $i$, $a_i>0$ so $a_i$ is positive for all hypernatural $i$. For all $a_j$ with $n\le j\le m$ the sum $\sum_m^na_i>a_j>0$, so if $\sum_m^na$ is infinitesimal,all $a_j$ are also infinitesimal.
The sequence $a_i/b_i$ converges to a limit $L$, so for all hypernaturals $j$ the term $a_j/b_j=L+\varepsilon_j$ where $\varepsilon$ is infinitesimal. So $a_j=b_j(L+\varepsilon_j)$. From this, if $L$ is non-zero $b_j$ must be infinitesimal.
Here, I tried rewriting the sum $\sum_m^n a_i=\sum_m^nb_i(L+\varepsilon_j)=L\sum_m^nb_i+\sum_m^nb_j\varepsilon_j$. If the rightmost sum is infinitesimal, then it would follow that $\sum_m^n b_j$ is infinitesimal.
The rightmost sum $\sum_m^nb_j\varepsilon_j$ is a sum of infinitesimals, so for all finite sums it is infinitesimal. However, if $n-m$ is unlimited, I don't know how to fix this argument.
Alas, this does not hold as stated. If you set $a_i = 2^{-i}$ and $b_i = i$, then $\frac{a_i}{b_i}$ converges to $0 \in \mathbb{R}$, and while $\sum_{\omega}^{\omega+1} a_i = 2^{-\omega}$ is infinitesimal, $\sum_{\omega}^{\omega+1} b_i = \omega$ very much isn't.
I think the author might have meant for $\frac{a_i}{b_i}$ to converge in $\mathbb{R}_+$. That, or else Goldblatt has some other condition in his definition of series that I failed to take into account - let me know if so, as I don't have my copy in hand to check.
How does one prove the positive limit case?
Consider two positive sequences $a$ and $b$ so that $\frac{a_k}{b_k} = L + \varepsilon_k$ for each unlimited $k$, where $L \in \mathbb{R}$, $L > 0$ and $\varepsilon_k$ is some infinitesimal depending on $k$.
Take unlimited $m < n$, and assume that $\sum_m^n b_i$ is infinitesimal. We wish to prove that $\sum_m^n a_i$ is infinitesimal. Since we know that for all unlimited $k$, $\frac{a_k}{b_k} = L + \varepsilon_k$, we can rearrange to get $a_k = L b_k + \varepsilon_k b_k$. Since $|\varepsilon_k| < 1$, we obtain the inequality
$$a_k = |a_k| = |L b_k + \varepsilon_k b_k| \leq L|b_k|+|\varepsilon_k b_k| < L|b_k| + |b_k|$$
which lets us write
$$\sum_m^n a_i = \sum_m^n |a_i| \leq L\sum_m^n \left|b_i\right| + \sum_m^n \left|b_i\right| = (L + 1) \sum_m^n b_i $$
The right hand side is a product of a real and an infinitesimal, so is itself infinitesimal. Then, since the left hand side is smaller than that infinitesimal and positive, it too must be infinitesimal.
Now, since $L \neq 0$ above, then $\frac{b_k}{a_k}$ is infinitesimally close to some positive real number as well, which gives the other implication (if $\sum_m^n a_i$ is infinitesimal then so is $\sum_m^n b_i$) by an identical argument.