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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 $$\lim_{n \to \infty}\dfrac{f(n)}{g(n)} = \dfrac{\lim_{n \to \infty}f(n)}{\lim_{n \to \infty}g(n)}.$$

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    Simply, you can't. For you to do that, the limits must first exist. This argument, however, does allow you to create asymptotes to a summation, show convergence/divergence, etc. – 2017-01-27
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    @SimplyBeautifulArt If the infinite sums are finite, I can? – 2017-01-27
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    By the definition of what it means to take an infinite sum and limit properties, yes, under the right conditions ofc. – 2017-01-27
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    @SimplyBeautifulArt What do you mean "under the right conditions"? – 2017-01-27
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    `I would think this result would hold because in limits we have` That only holds if $g(n) \ne 0$ and $\lim g(n) \ne 0$. As long as the partial sums satisfy those conditions, and $\lim f(n)$ exists, the limits do indeed permute. – 2017-01-27
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    Basic conditions like $\lim_{n\to\infty}f(n)$ existing and what-not. – 2017-01-27
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    The question is no different from asking whether $\lim_{n\to \infty}\frac{a_n}{b_n}=\frac{\lim_{n\to \infty}a_n}{\lim_{n\to \infty}b_n}$. The answer is yes if the individual limits exist and if the limit of $b_n\ne 0$. The answer is not in general and an easy counter example is $a_n=b_n=n$ – 2017-01-27

2 Answers 2

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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 $$ \lim_{m \to \infty}\dfrac{\sum_{n=0}^{m}f(n)}{\sum_{n=0}^{m}g(n)}=\frac{a}{b} $$

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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.