Does $\frac{\sum_{n=0}^{\infty}f(n)}{\sum_{n=0}^{\infty}g(n)} = \lim\limits_{m \to \infty}\frac{\sum_{n=0}^{m}f(n)}{\sum_{n=0}^{m}g(n)}$? 
Does $\dfrac{\sum_{n=0}^{\infty}f(n)}{\sum_{n=0}^{\infty}g(n)} = \lim_{m \to \infty}\dfrac{\sum_{n=0}^{m}f(n)}{\sum_{n=0}^{m}g(n)}$ where $f(n)$ and $g(n)$ are functions in $n$ if the infinite sums are finite and nonzero?

Otherwise how could we make sense of sums such as $\dfrac{\sum_{n=0}^{\infty}n}{\sum_{n=0}^{\infty}\dfrac{1}{2}n}$ since the sums in the numerator and denominator both go to infinity. Also does this result hold if the sums in the numerator and denominator are finite?
I would think this result would hold because in limits we have $$\lim_{n \to \infty}\dfrac{f(n)}{g(n)} = \dfrac{\lim_{n \to \infty}f(n)}{\lim_{n \to \infty}g(n)}.$$
 A: Assume 
$$
\lim_{m \to \infty}\sum_{n=0}^{m}f(n)=\sum_{n=0}^{\infty}f(n)=a\quad\text{and}\quad\lim_{m \to \infty}\sum_{n=0}^{m}g(n)=\sum_{n=0}^{\infty}g(n)=b\ne 0
$$
We can prove that
$$
\lim_{m \to \infty}\dfrac{\sum_{n=0}^{m}f(n)}{\sum_{n=0}^{m}g(n)}=\dfrac{\sum_{n=0}^{\infty}f(n)}{\sum_{n=0}^{\infty}g(n)}=\frac{a}{b}
$$
By definition, for any $\epsilon>0$, there is a $N_1$ such that for any $m>N_1$
$$
\left|\sum_{n=0}^{m}f(n)-a\right|<\epsilon
$$
And a $N_2$ such that for any $m>N_2$
$$
\left|\sum_{n=0}^{m}g(n)-b\right|<\epsilon
$$
Let $\epsilon=|b|/2$. Then there is a $N_3$ such that for any $m>N_3$
$$
\left|\sum_{n=0}^{m}g(n)\right|>\frac{|b|}{2}
$$
Let $N=\max{(N_1, N_2, N_3)}$. Since for any $m>N$, there is
\begin{align}
\left|\dfrac{\sum_{n=0}^{m}f(n)}{\sum_{n=0}^{m}g(n)}-\frac{a}{b}\right|&=\left|\dfrac{b\sum_{n=0}^{m}f(n)-a\sum_{n=0}^{m}g(n)}{b\sum_{n=0}^{m}g(n)}\right|
\\
&<\frac{2}{b^2}\left|b\sum_{n=0}^{m}f(n)-ab+ab-a\sum_{n=0}^{m}g(n)\right|
\\
&=\frac{2}{b^2}\left|b\left(\sum_{n=0}^{m}f(n)-a\right)-a\left(\sum_{n=0}^{m}g(n)-b\right)\right|
\\
&<\frac{2}{b^2}(|b|+|a|)\epsilon
\end{align}
We have
$$
\lim_{m \to \infty}\dfrac{\sum_{n=0}^{m}f(n)}{\sum_{n=0}^{m}g(n)}=\frac{a}{b}
$$
A: Ok, here's how you could do it:
Assume that $\sum_{n\geq0}f(n)=a$ and $\sum_{n\geq0}g(n)=b\neq0$. We will show that $\frac{\sum_{n=0}^m f(n)}{\sum_{n=0}^mg(n)}\rightarrow \frac ab$ as $m\rightarrow\infty$ using the usual definition of convergence.
Fix $\varepsilon>0$ and choose M so large that for all $k\geq M$, $$\left\lvert \sum_{n\geq k}f(n)\right\rvert,\left\lvert \sum_{n\geq k}g(n)\right\rvert\leq\min\left\{\frac{b^2}{2(\lvert a\rvert+\lvert b\rvert)}\varepsilon,\frac{\lvert b\rvert}{2}\right\}.$$
Now, if $k\geq M$, we have
$$\left\lvert\frac{\sum_{n\leq k}f(n)}{\sum_{n\leq k}g(n)}-\frac ab \right\rvert=\left\lvert \frac{b \sum_{n\leq k}f(n)-a\sum_{n\leq k}g(n)}{b \sum_{n\leq k}g(n)} \right\rvert\leq \left\lvert \frac{b \sum_{n\leq k}f(n)-a\sum_{n\leq k}g(n)}{b\frac { b}{2}}\right\rvert$$
For the nominator, we have:
$$b \sum_{n\leq k}f(n)-a\sum_{n\leq k}g(n)=(b(a-\sum_{n>k} f(n))-a(b-\sum_{n>k}g(n))\\=-b\sum_{n>k}f(n)+a\sum_{n>k}g(n),$$
so that by taking absolute value, we get (upon using the triangle inequality)
$$\left\lvert -b\sum_{n>k}f(n)+a\sum_{n>k}g(n)\right\rvert\leq (\lvert a\rvert+\lvert b\rvert)\frac{b^2}{2(\lvert a\rvert+\lvert b\rvert)}\varepsilon=\frac{b^2}{2}\varepsilon .$$
By plugging this back into the top inequality  and as $\varepsilon>0$ was arbitrary, the claim is indeed proved.
