$\lim \frac{a_{2n}}{a_n}< \frac{1}{2}$ 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.
 A: 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.
A: 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.
