why does $2^{-n}n^{1000} $ converge by the limit comparison test? the series $\displaystyle\sum\limits_{n=1}^{\infty}  2^{-n}n^{1000}$ converges by the comparison test. 
$2^{-n}\leqslant \displaystyle\frac{c}{n^{1002}}$
$\displaystyle\sum\limits_{n=1}^{\infty}  2^{-n}n^{1000} \leqslant \displaystyle\sum\limits_{n=1}^{\infty}n^{1000}\frac{c}{n^{1002}} =c\displaystyle\sum\limits_{n=1}^{\infty}\frac{1}{n^{2}} $
$2 \geqslant 1$ converges by the p-test thus $\displaystyle\sum\limits_{n=1}^{\infty}  2^{-n}n^{1000}$ converges by the comparison test. 
my question is, what is to prevent someone from choosing $\displaystyle\frac{c}{n^{1001}}$ which would make the series harmonic thus divergent instead of convergent. I don't feel like it is a valid proof for the series. maybe there is something i am missing here, can someon explain to me why this is a valid proof for the convergences of the series. 

Because for large $n$,  $2^{-n}$ decreases rapidly as compared to $\frac{1}{n^p}$ for any $p>0$. This you can prove by using Binomial theorem. So there exists $n_0$ and $c$ such that $$2^{-n}<\frac{c}{n^p}, ~~\text{for } n>n_0.$$
Now if $a_n\leq b_n$, and if $\sum b_n$ is divergent, by Comparison test nothing can be derived about convergence/divergence of  $\sum a_n$. Hence for $p=1001$ nothing can be derived, but for $p=1002$ Comparison test works.  
