I'm having a little trouble solving this problem :
Consider two sequences of real numbers, $\ (a_n)_{n \ge 0}$, and $\ (b_n)_{n \ge 0}$, with every $\ b_n >0 $, such that $\frac{a_n}{b_n} $ converges to a finite limit S.
Show that $\frac{\sum_{n=0}^{N}{a_n} }{\sum_{n=0}^{N}{b_n} } $ converges to that same limit S, when N approaches $ +\infty$.
The way I'm approaching this problem is by reasoning with the epsilon definition of a limit :
Consider $ \epsilon >0 $. There exists $ N_{0}\in \mathbb N $ such that, for every $ n\ge N_{0}$ : $ \left| \frac{a_n}{b_n} -S \right| \le \epsilon $
In that case, I would think it useful to write :
For every $ N\ge N_{0}$ : $ \left| \frac{\sum_{n=0}^{N}{a_n} }{\sum_{n=0}^{N}{b_n} } -S \right| = \left| \frac{\sum_{n=0}^{N}{a_n} }{\sum_{n=0}^{N}{b_n} } -\frac{a_N}{b_N} + \frac{a_N}{b_N} -S \right| \le \left| \frac{\sum_{n=0}^{N}{a_n} }{\sum_{n=0}^{N}{b_n} } -\frac{a_N}{b_N} \right|+\left| \frac{a_N}{b_N}-S \right| \le \epsilon+ \left| \frac{\sum_{n=0}^{N}{a_n} }{\sum_{n=0}^{N}{b_n} } -\frac{a_N}{b_N} \right| $
Another useful strategy might be to find lower and upper bounds of $\frac{\sum_{n=0}^{N}{a_n} }{\sum_{n=0}^{N}{b_n} } $ that both converge to .
Unfortunately, I've hit a roadblock on both these methods, and don't really see how to go forward.
Any help would be appreciated, thanks.
It's wrong as stated. Take $a_n=\frac{1}{n^2\log n}$ and $b_n = \frac{1}{n^2}$. Then $\frac{a_n}{b_n} = \frac{1}{\log n} \to 0$ but $\frac{\sum_{n=1}^N a_n}{\sum_{n=1}^N b_n} \to_{N \to \infty} c \neq 0$.