# $\lim \frac{a_{2n}}{a_n}< \frac{1}{2}$ series convergent

Could you tell me how to prove such lemmas?

We are given a decreasing sequence of positive numbers $(a_n)$

1) If $\lim_{n \rightarrow \infty} \frac{a_{2n}}{a_n} =g< \frac{1}{2}$ then the series $\sum _{n=1} ^{\infty} a_n$ is convergent

2)If $\lim_{n \rightarrow \infty} \frac{a_{2n}}{a_n} =G > \frac{1}{2}$ then the series $\sum _{n=1} ^{\infty} a_n$ is divergent

I know it must be very easy but I don't know what to do about $2n$ in $a_{2n}$ (which I gather is the main issue of the whole proof).

Do you think you could help me?

Thank you.

-
I assume you are talking about the convergence of $\sum a_n$. Start with Cauchy Condensation. Your problem will be a quick consequence. If the limit is $\lt 1/2$, then there is $\alpha\lt 1/2$ such that after a while the ratio is $\lt \alpha$. Or if you wish, instead of using Cauchy Condensation, you can imitate the proof. –  André Nicolas Feb 26 '13 at 18:31
He says "the series is convergent" which presumably means $\sum a_i$. But he is vague on the point. @Maesumi –  Thomas Andrews Feb 26 '13 at 18:34
Yes, you are right. I've already corrected my question. –  Hagrid Feb 26 '13 at 18:37

Look up the Cauchy Condensation Test. Then either use the test, or adapt the proof.

Because it is faster, we use the test. For example, for the first part, there is an $\alpha \lt \frac{1}{2}$ and an $N$ such that if $n \ge N$, then $\frac{a_{2n}}{a_n} \lt \alpha$.

But by Cauchy Condensation, $\sum a_k$ converges if and only if $\sum 2^k a_{2^k}$ converges. The latter series converges by long run comparison with the geometric series $\sum (2\alpha)^k$.

Remark: Adapting the proof instead of using the result is a very good idea. One can then see that Condensation, which at first appears magical, comes from natural estimates.

-
I've already analysed the proof of the Cauchy Condensation Test. But could you explain a little bit further how to use it for $\frac{a_{2n}}{a_n}$ ? –  Hagrid Feb 26 '13 at 19:45
How to use the Test? Let $n$ be a large power of $2$, say $2^M$, and let $b=a_n$. Then for $j \ge M$, $a_{2^j}\lt b\alpha^{j-M}=b\alpha^{-M}\alpha^j$. Multiplying by $2^j$ gives us $\lt b\alpha^{-M}(2\alpha)^j$, the terms of a convergent geometric series. –  André Nicolas Feb 26 '13 at 19:53
Thank you a lot for your help. It's all clear now. –  Hagrid Feb 26 '13 at 20:08

Denote the partial sums by $S_n$. Now you can write \begin{align} \sum_i a_i = S &= S_n + (S_{2n}-S_{n}) + (S_{4n}-S_{2n}) + (S_{4n}-S_{2n})+\cdots \\ &= S_n + T_1 + T_2 + T_3+\cdots \\ \end{align} To show that the series is convergent you easily get lower bounds of the form $c (2g)^k$ for $T_k$ in case 1) and upper bounds of the form $c (2G)^k$ for $T_k$ in case 2). (You have to make use of the fact that the $a_i$ are decreasing, too). So, in case one you've bounded your series with a convergent geometric series from above, and in case two with a divergent geometric series from below.

EDIT: Ok, I saw that this is pretty similar to the proof of the Cauchy condensation test, that has been mentioned by Andre already. I'll still leave it in here.

-