Convergence of $\sum_{n} \frac{a_n}{2^n}$ where $a_n\rightarrow +\infty$ Does there exists a sequence $a_n$ of positive reals such that $a_n\rightarrow \infty$ but $\sum_{n}\frac{a_n}{2^n}$ converges? (I considered $a_n=n$, but couldn't prove convergence or divergence of $\sum_{n}\frac{a_n}{2^n}$.)
 A: Start here:
$$ \frac{1}{1-x}=\lim_{n\rightarrow\infty}\sum_{k=0}^{n} x^k$$
I will omit the limit from now, but know that it should be there. Replace x by $\frac{x}{2}$ to get
$$ \frac{1}{1-\frac{x}{2}}=\sum_{k=0}^{n} \left(\frac{x}{2}\right)^k$$
simplify on the left
$$ \frac{2}{2-x}=\sum_{k=0}^{n} \left(\frac{x}{2}\right)^k$$
now take the derivative on both sides,
$$ \frac{2}{(2-x)^2}=\sum_{k=1}^{n} kx^{k-1}\left(\frac{1}{2}\right)^{k}$$
let $x=1$ to get 
$$ \frac{2}{(2-1)^2}=\sum_{k=1}^{n} k\left(\frac{1}{2}\right)^{k}$$
$$ 2=\sum_{k=1}^{n} \frac{k}{2^{k}}$$
and voilà $a_n=n$ works. 
A: Any $a_n$
such that
$\lim \sup (a_n)^{1/n}
< 2$.
Proof:
If 
$\lim \sup (a_n)^{1/n}
< 2$,
there is an real $c$
such that
$2 >c>0$
and a positive integer $m$
such that,
for $n > m$,
$(a_n)^{1/n} < 2-c$
or
$a_n
< (2-c)^n
$.
Therefore,
using the traditional splitting of the sum
into two parts,
$\begin{array}\\
\sum_{n=0}^N \frac{a_n}{2^n}
&=\sum_{n=0}^m \frac{a_n}{2^n}+\sum_{n=m+1}^N \frac{a_n}{2^n}\\
&\le (m+1)\max_{n=0}^m \frac{a_n}{2^n}+\sum_{n=m+1}^N \frac{a_n}{2^n}\\
&<(m+1)d_m+\sum_{n=m+1}^N \frac{(2-c)^n}{2^n}\\
&=(m+1)d_m+\sum_{n=m+1}^N (1-c/2)^n\\
\end{array}
$
and both terms are bounded
so
$\sum_{n=0}^N \frac{a_n}{2^n}$
is a bounded and increasing sequence,
and, therefore,
converges.
Examples of 
such sequences are
$(3/2)^n$,
$(b^n)$
for $0 < b < 2$,
and
$(n^k)$
for and $k > 0$.
