How can we conclude that this series converges faster?

Question

let $u_k = 1- L \frac{b_k}{a_k}$ where both $\sum a_k$ and $\sum b_k$ are convergent and $$\lim \frac{a_n}{b_n}=L.$$

Now I am comparing $\sum a_k$ and $\sum u_ka_k$

Textbook tells me :

since $u_k \to 1- L{\frac{1}{L}}$ It approaches to its value faster than the series $\sum a_k$

I couldn't get how? I mean $u_k$ is a sequence, not a series. How can we conclude if the sequence approaches 0 faster that the series approaches its convergent value faster? How do we do that? And How come did we come up with this statement anyway? I can compare two series that I know in case of fastness etc (i.e $\sum \frac{1}{n}$ and $\sum \frac{1}{n^2}$) but in this question I'm not given $a_k$

Answer 1

To compare the speed of convergence, we look at the remainder sums

$$\sum_{k = n}^{\infty} a_k$$

and analogously for the series with terms $u_ka_k$. Per your comment, everything is positive, so by definition, we have $u_k \to 0$ (for this we need $L\neq 0$, it wouldn't matter if we had $L < 0$). Thus, for every $\varepsilon > 0$ there is an $N(\varepsilon)\in \mathbb{N}$ such that $\lvert u_k\rvert \leqslant \varepsilon$ for $k \geqslant N(\varepsilon)$. Then

$$\Biggl\lvert \sum_{k = n}^m u_ka_k\Biggr\rvert \leqslant \sum_{k = n}^m \lvert u_ka_k\rvert = \sum_{k = n}^m \lvert u_k\rvert a_k \leqslant \varepsilon \sum_{k = n}^m a_k$$

for all $N(\varepsilon) \leqslant n \leqslant m$, and taking the limit $m \to \infty$ we obtain

$$\Biggl\lvert \sum_{k = n}^\infty u_ka_k\Biggr\rvert \leqslant \varepsilon \sum_{k = n}^{\infty} a_k.$$

Setting

$$A := \sum_{k = 0}^\infty a_k\quad\text{and}\quad U := \sum_{k = 0}^{\infty} u_ka_k$$

we have

$$\Biggl\lvert U - \sum_{k = 0}^{n-1} u_ka_k\Biggr\rvert \leqslant \varepsilon\Biggl(A - \sum_{k = 0}^{n-1} a_k\Biggr)$$

for $n \geqslant N(\varepsilon)$. That is,

$$\lim_{n\to \infty} \frac{\Bigl\lvert U -\sum_{k = 0}^{n} u_ka_k\Bigr\rvert}{A - \sum_{k = 0}^{n} a_k} = 0.$$