constructing a sequence that converges to zero fast enough (for the corresponding series to converge) Suppose there is a sequence in $\mathbb{R}$ that converges to $0$, but such that the corresponding series does not necessarily converge.  In this general case, how can I construct a sequence that is guaranteed to produce a convergent series using the terms of the original one?  
I know if my example is $X_n = \{\frac{1}{n}\}$ then I could just take the terms that are powers of $\frac{1}{2}$ from $X_n$ and construct a new sequence such that when summed converges (since it is the geometric series).  
How can I do this in the very general case 
 A: Start by picking your favorite convergent positive series $\sum a_n$. You can take $a_n=\frac{1}{2^n}$ for example. Now let $X_n$ be a sequence that converges to $0$. Choose a subsequence $X_{n_k}$ such that for every $k$ you have$$|X_{n_k}|<a_k.$$This guarantees that $\sum X_{n_k}$ is absolutely convergent and in particular convergent.
A: Take the elements $n_k$ such that $n_k\sim k^{1+\epsilon}$ where $\epsilon>0$ i.e $lim_k \frac{n_k}{k^{1+\epsilon}}=1$
A: You can use your idea more generally. You know that
$$\sum_{n=1}^\infty \frac{1}{2^n} = \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots$$converges and you have a sequence $(a_n)$ with (I assume) positive terms, converging to $0$. You want to construct a sequence $b_n$ using terms from the original sequence but with $\sum b_n$ convergent.
Since $a_n\to 0$, you can find $i_1 \in \mathbb{N}$ such that $a_{i_1} < \tfrac{1}{2}$, the first $i_1$ that works will do: set $b_1=a_{i_1}$. For the next term, take $i_2 > i_1$ such that $a_{i_2} < \tfrac{1}{4}$ and set $b_2 = a_{i_2}$. Continue by finding $i_k$ such that $a_{i_k} < \tfrac{1}{2^k}$ and set $b_k = a_{i_k}$.
In this way, you construct a sequence with terms $b_n$ taken from the $a_n$'s but in such a way that you have $b_n < \tfrac{1}{2^n}$ for all $n$, so $\sum b_n$ converges by the comparison test.
