I've got an interesting solution I think, it uses partial sums of diverging series with positive terms:
Let: $b_n = \frac{2^n}{n}$ ; $v_n = b_{n}-b_{n-1}$
$ v_n = \frac{2^n}{n}-\frac{2^{n-1}}{n-1} = \frac{2^{n-1}}{n-1}*[ 1-\frac{2}{n} ] $ ~ $ \frac{2^{n-1}}{n-1} = b_{n-1}$ , when $ n \rightarrow +\infty $
$(b_n) \rightarrow +\infty $, when $n \rightarrow +\infty$ , so $\sum v_n$ diverges and we can say that both partial sums are equivalent:
$ \sum_{k=2}^{n+1} v_k = b_{n+1}-b_1 $ ~ $b_{n+1}$ ~ $ \sum_{k=1}^n b_k $ , when $ n \rightarrow +\infty $
Hence you get: $ a_n = \frac{1}{b_{n+1}}*\sum_{k=1}^n b_k $ ~ 1 , when $ n \rightarrow +\infty $
You get the limit :) , which is 1 like you said
Edit: I'll prove the property I used above. Let's have $ \sum u_n$ and $\sum v_n $ two diverging series of positive terms: $ u_n,v_n \geq 0 $ such as: $u_n$ ~ $v_n$
$u_n$ ~ $v_n$ <=> $ u_n = v_n + o(v_n) $
$ u_n = v_n + o(v_n) $ <=> for any $ \epsilon > 0 , n \geq N => u_n -v_n \leq \frac{\epsilon}{2}*v_n $
Now we sum this from N+1 to n :
$ \sum_{k =N+1}^n u_k -\sum_{k =N+1}^n v_k \leq \frac{\epsilon}{2}*\sum_{k =N+1}^n v_k $
=> $ \sum_{k =1}^n u_k -\sum_{k =1}^n v_k \leq \frac{\epsilon}{2}*\sum_{k =1}^n v_k + \sum_{k =1}^N u_k -(\frac{\epsilon}{2} +1)*\sum_{k =1}^N v_k $
=> $ \sum_{k =1}^n u_k -\sum_{k =1}^n v_k \leq \frac{\epsilon}{2}*\sum_{k =1}^n v_k + \sum_{k =1}^N u_k $ , since $ (\frac{\epsilon}{2} +1)*\sum_{k =1}^N v_k \geq 0 $
Now I'll define : $U_n = \sum_{k =1}^n u_k$, $ V_n = \sum_{k =1}^n v_k $
The hypothesis is : $(V_n) \rightarrow +\infty$ ; so : $ \sum_{k =1}^N u_k =o(V_n) $
=> $ \epsilon > 0 , n \geq N_o => \sum_{k =1}^N u_k \leq \frac{\epsilon}{2}*\sum_{k =1}^n v_k $
You get : $ \epsilon >0 , n > max(N, N_o) => U_n -V_n \leq \epsilon*V_n $
That's exactly : $ U_n -V_n = o(V_n) $ ie $U_n$ ~ $V_n$