Convergence of a sequence implies the convergence of another sequence 
Let $a_n>0$, $n≥ 0$. Call $k$ “good” if $k\geqslant 1$ and $a_k>\frac12 a_{k-1}$. Also call $0$ good. Show $\sum^{\infty}_{k=0}a_k$ converges if $\sum\limits_{k \text{ is good }} a_k$ converges. 

I can only see that the sum of the good $k$ terms is greater than half of their previous terms sum, how to relate the good $k$ terms with every term of this sequence? 
 A: If after a while all $k$ are good, then the result is clear. For the convergence of a series is not affected by prepending a finite number of terms.
If after a while all $k$ are bad, then again the result is clear, by comparison with a geometric series.
Now we deal with the case where there are infinitely many good and infinitely many bad. 
We need not worry about  the terms of the sequence before the first good index, since they do not affect convergence. 
The idea of the proof is to show that the bad $k$ cannot cause trouble, the sum of the $a_i$ for bad $i$ is "small." 
Let $k$ be any bad index. Let $m$ be the largest good index which is less than $k$, and let $a_n$ be the smallest good index which is greater than $k$.  
Then $a_{m+1}\le \frac{1}{2}a_m$, and $a_{m+2}\le \frac{1}{4}a_m$, and so on. Thus the sum of the $a_i$ such that $i$ lies in the interval $m\lt i\lt n$ is $\lt {a_m}$. 
Thus the sum of all $a_j$ such that $j$ is bad and lies between two goods is $\lt$ the sum of all $a_k$ with $k$ good. 
Thus the sequence of partial sums of our full series is bounded, and therefore the series converges. 
A: Notice that
$$\mathbb N=\underbrace{\{k\ |\ \text{k is good}\}}_{=G}\cup \underbrace{\{k\ |\ \text{k isn't good}\}}_{=B}$$
so to find the result it suffices to show that
$$\sum\limits_{k \text{ isn't good }} a_k$$
is convergent. 
If $G$ is empty set the result is  trivial, otherwise Let  $k\in G$ such that $k+1\in B$ then $$a_{k+1}\leq \frac{1}{2}a_k$$
and if $k+2,k+3,\ldots,k+s\in B$ and $k+s+1\in G$ then
$$a_{k+2} \leq\frac{1}{2}a_{k+1}\leq \frac{a_k}{2^2}$$
$$\vdots$$
$$a_{k+s} \leq \frac{a_k}{2^s}$$
so
$$\sum_{p=1}^sa_{k+p}\leq \sum_{p=1}^s \frac{a_k}{2^p}\leq \sum_{p=1}^\infty \frac{a_k}{2^p}=a_k$$
hence we can see that
$$\sum\limits_{k \text{ isn't good }} a_k\leq \sum\limits_{k \text{ is good }} a_k$$
and the result follows easily.
A: Given the sequence $(a_k)_k$, let $(k_m)_m
$ be the unique strictly increasing sequence of "good" integers.
Suppose $\sum^{\infty}_{m=1}a_{k_m}$ converges to $l$.
Since $a_k>\frac12 a_{k-1}$ for all $k$ ,we have the following:
$\sum^{\infty}_{k=0}a_k$ = $\sum^{k={k_1}-1}_{k=0}a_{k} + \sum^{k={k_2}-1}_{k={k_1}}a_{k} + \sum^{k={k_3}-1}_{k={k_2}}a_{k} + ...$
$\leq  \sum^{k={k_1}-1}_{k=0}a_{k} + 2 a_{{k_1}} + 2 a_{{k_2}} + 2 a_{{k_3}} + ...$
$ =\sum^{k={k_1}-1}_{k=0}a_{k} + 2\sum^{\infty}_{m=1}a_{k_m}$
$ =\sum^{k={k_1}-1}_{k=0}a_{k} + 2l$
