# 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.

• If you chose $c/n^{10001}$ you would bound the series from above by a divergent series. This would imply that the sum, roughly, is less than or equal to infinity. This tells you nothing. The comparison test is inconclusive with this choice. Commented Oct 26, 2017 at 2:13
• You have some arithmetic mistakes. Correctly: $$\sum_{n=1}^\infty n^{1000}\frac c{n^{10002}}=c\sum_{n=1}^\infty\frac1{n^{9002}}$$ $$\sum_{n=1}^\infty n^{1000}\frac c{n^{10001}}=c\sum_{n=1}^\infty\frac1{n^{9001}}$$
– bof
Commented Oct 26, 2017 at 6:08
• yeah, i didn't mean to put that extra 0
– Катя
Commented Oct 27, 2017 at 0:21

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.