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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.

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    $\begingroup$ 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. $\endgroup$ Commented Oct 26, 2017 at 2:13
  • $\begingroup$ You have some arithmetic mistakes. Correctly: $$\sum_{n=1}^\infty n^{1000}\frac c{n^{10002}}=c\sum_{n=1}^\infty\frac1{n^{9002}}$$ $$\sum_{n=1}^\infty n^{1000}\frac c{n^{10001}}=c\sum_{n=1}^\infty\frac1{n^{9001}}$$ $\endgroup$
    – bof
    Commented Oct 26, 2017 at 6:08
  • $\begingroup$ yeah, i didn't mean to put that extra 0 $\endgroup$
    – Катя
    Commented Oct 27, 2017 at 0:21

1 Answer 1

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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.

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