# Show convergence/divergence of series

We have that $(a_k)$ and $(b_k)$ are two sequences of positive numbers.

I want to show the following:

1. If $\lim_{k\rightarrow\infty}\frac{a_k}{ b_k}= c > 0$, then both $\sum_{k=1}^{\infty}a_k$ and $\sum_{k=1}^{\infty}b_k$ converge or both diverge.

2. If $\frac{a_k}{ b_k}\geq \frac{a_{k+1}}{ b_{k+1}}$ for almost each $k$, then from the convergence of $\sum_{k=1}^{\infty}b_k$ we get the convergence of $\sum_{k=1}^{\infty}a_k$ and from the divergence of $\sum_{k=1}^{\infty}a_k$ we get the divergence of $\sum_{k=1}^{\infty}b_k$.



1. Could you give me a hint how we could show that? Do we have to apply a convergence test?

2. We have that $$\frac{a_k}{ b_k}\geq \frac{a_{k+1}}{ b_{k+1}}\Rightarrow \frac{b_{k+1}}{ b_k}\geq \frac{a_{k+1}}{ a_k}$$ Do we apply here the ration and the direct comparison test?

For the first one, if $\lim_{k \to \infty} \frac{a_k}{b_k} = c$, then note that if we take $\epsilon = \frac c2$ in the definition, then for some large enough $N$, if $n > N$ then $\frac c2 < \frac{a_k}{b_k} \leq \frac{3c}2$. This means, that for any $l > 0$, $$\frac c2 \sum_{k=N}^{N+l} b_k \leq \sum_{k=N}^{N+l}a_k \leq \frac{3c}{2}\sum_{k=N}^{N+l} b_k$$

Note that the first few terms $a_1,...,a_N$ and $b_1,...,b_N$ don't affect divergence or convergence, so we neglect them.

Now, suppose that $\sum_{n=N}^\infty a_k$ diverges. By comparison, $\frac {3c}2 \sum_{k=N}^\infty b_k$ also diverges, but then $\frac {3c}{2}$ is just a constant, so $\sum b_k$ diverges. Of course, if $\sum a_k$ converges, then from above, $\sum b_k$ converges by the comparison test.

I presume, that by "almost each", you mean that for some large enough $N$, it will always be true that $\frac{a_k}{b_k}$ is decreasing with $k$ for $k>N$. In that case, take $d = \frac{a_N}{b_N}$ and note that $\sum_{k=N}^{N+l} a_k \leq d \sum_{k=N}^{N+l} b_k$.

Now, both statements made in the second question follow from the above inequality : if $b_n$ converges, so does $a_n$ by comparison, and if $a_n$ diverges, so does $b_n$ by comparison.

• We have that $\frac{a_k}{b_k}$ is decreasing and that $k>N$. So we get that $\frac{a_k}{b_k}<\frac{a_N}{b_N}$. Do we take then the limit of this inequality? Or why can we take $c = \frac{a_N}{b_N}$ ? Or isn't this symbol $c$ as in the first question? – Mary Star Dec 2 '17 at 23:45
• I've just renamed $c$ as $d$ to avoid confusion with the first question. See, it's very simple : if $\frac{a_k}{b_k}$ is decreasing with $k$ for $k > N$, then $a_k < db_k$ for every $k > N$. So, you can now sum over $k=N$ to $N+l$ on both sides, because inequality is preserved by addition. Then, you can use the comparison test. – астон вілла олоф мэллбэрг Dec 2 '17 at 23:48
• Ah ok. And to get the infite sum we take $l\rightarrow \infty$, or not? – Mary Star Dec 2 '17 at 23:57
• Correct. Since the inequality is true for any $l$, we take $l \to \infty$ to get the infinite sum. – астон вілла олоф мэллбэрг Dec 2 '17 at 23:57
• You are welcome. Thank you for your interaction, and $+1$ to your question. – астон вілла олоф мэллбэрг Dec 3 '17 at 0:02