How to show divergence of $sum_{n = 1}^{infty} 2^{-o(1)}frac{1}{n}$?
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I am trying to show the divergence of the following sum:
$$sum_{n = 1}^{infty} 2^{-o(1)}frac{1}{n}$$
where the exponent of the 2 is negative little-o of 1. I have tried some ideas such as the following:
since $o(1)$ means that $lim_{n rightarrow infty} f(n)/1 rightarrow 0$, then we can potentially look at $max_n f(n)$ and $min_n f(n)$ and use those to bound the term $2^{-o(1)}$ in the sum, but I'm not sure how to handle functions $f(n)$ that oscillate. Any thoughts are appreciated!
sequences-and-series analysis computational-complexity
asked Dec 10 '18 at 21:43
hologramhologram
14312
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add a comment |
$begingroup$
I am trying to show the divergence of the following sum:
$$sum_{n = 1}^{infty} 2^{-o(1)}frac{1}{n}$$
where the exponent of the 2 is negative little-o of 1. I have tried some ideas such as the following:
since $o(1)$ means that $lim_{n rightarrow infty} f(n)/1 rightarrow 0$, then we can potentially look at $max_n f(n)$ and $min_n f(n)$ and use those to bound the term $2^{-o(1)}$ in the sum, but I'm not sure how to handle functions $f(n)$ that oscillate. Any thoughts are appreciated!
sequences-and-series analysis computational-complexity
asked Dec 10 '18 at 21:43
hologramhologram
14312
$endgroup$
$begingroup$
An $o(1)$ function cannot oscillate wildly.
$endgroup$
– Yves Daoust
Dec 10 '18 at 21:56
$begingroup$
apologies I have fixed that in the question
$endgroup$
– hologram
Dec 10 '18 at 21:59
add a comment |
$begingroup$
I am trying to show the divergence of the following sum:
$$sum_{n = 1}^{infty} 2^{-o(1)}frac{1}{n}$$
where the exponent of the 2 is negative little-o of 1. I have tried some ideas such as the following:
since $o(1)$ means that $lim_{n rightarrow infty} f(n)/1 rightarrow 0$, then we can potentially look at $max_n f(n)$ and $min_n f(n)$ and use those to bound the term $2^{-o(1)}$ in the sum, but I'm not sure how to handle functions $f(n)$ that oscillate. Any thoughts are appreciated!
sequences-and-series analysis computational-complexity
asked Dec 10 '18 at 21:43
hologramhologram
14312
$endgroup$
I am trying to show the divergence of the following sum:
$$sum_{n = 1}^{infty} 2^{-o(1)}frac{1}{n}$$
where the exponent of the 2 is negative little-o of 1. I have tried some ideas such as the following:
since $o(1)$ means that $lim_{n rightarrow infty} f(n)/1 rightarrow 0$, then we can potentially look at $max_n f(n)$ and $min_n f(n)$ and use those to bound the term $2^{-o(1)}$ in the sum, but I'm not sure how to handle functions $f(n)$ that oscillate. Any thoughts are appreciated!
sequences-and-series analysis computational-complexity
sequences-and-series analysis computational-complexity
asked Dec 10 '18 at 21:43
hologramhologram
14312
asked Dec 10 '18 at 21:43
hologramhologram
14312
edited Dec 10 '18 at 21:58
hologram
asked Dec 10 '18 at 21:43
hologramhologram
14312
asked Dec 10 '18 at 21:43
hologramhologram
14312
asked Dec 10 '18 at 21:43
hologramhologram
14312
14312
$begingroup$
An $o(1)$ function cannot oscillate wildly.
$endgroup$
– Yves Daoust
Dec 10 '18 at 21:56
$begingroup$
apologies I have fixed that in the question
$endgroup$
– hologram
Dec 10 '18 at 21:59
add a comment |
$begingroup$
An $o(1)$ function cannot oscillate wildly.
$endgroup$
– Yves Daoust
Dec 10 '18 at 21:56
$begingroup$
apologies I have fixed that in the question
$endgroup$
– hologram
Dec 10 '18 at 21:59
$begingroup$
An $o(1)$ function cannot oscillate wildly.
$endgroup$
– Yves Daoust
Dec 10 '18 at 21:56
$begingroup$
An $o(1)$ function cannot oscillate wildly.
$endgroup$
– Yves Daoust
Dec 10 '18 at 21:56
$begingroup$
apologies I have fixed that in the question
$endgroup$
– hologram
Dec 10 '18 at 21:59
$begingroup$
apologies I have fixed that in the question
$endgroup$
– hologram
Dec 10 '18 at 21:59
add a comment |
1 Answer
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$begingroup$
If $f(n)=o(1)$, by definition, for some $epsilon>0$, you will find an $N$ such that
$$n>Nimplies|f(n)|<epsilonimplies 2^{-f(n)}>2^{-epsilon}.$$
Then the tail of the sum diverges as it is bounded below by an harmonic series.
answered Dec 10 '18 at 21:55
Yves DaoustYves Daoust
127k673226
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1 Answer
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active
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
$begingroup$
If $f(n)=o(1)$, by definition, for some $epsilon>0$, you will find an $N$ such that
$$n>Nimplies|f(n)|<epsilonimplies 2^{-f(n)}>2^{-epsilon}.$$
Then the tail of the sum diverges as it is bounded below by an harmonic series.
answered Dec 10 '18 at 21:55
Yves DaoustYves Daoust
127k673226
$endgroup$
add a comment |
$begingroup$
If $f(n)=o(1)$, by definition, for some $epsilon>0$, you will find an $N$ such that
$$n>Nimplies|f(n)|<epsilonimplies 2^{-f(n)}>2^{-epsilon}.$$
Then the tail of the sum diverges as it is bounded below by an harmonic series.
answered Dec 10 '18 at 21:55
Yves DaoustYves Daoust
127k673226
$endgroup$
add a comment |
$begingroup$
If $f(n)=o(1)$, by definition, for some $epsilon>0$, you will find an $N$ such that
$$n>Nimplies|f(n)|<epsilonimplies 2^{-f(n)}>2^{-epsilon}.$$
Then the tail of the sum diverges as it is bounded below by an harmonic series.
answered Dec 10 '18 at 21:55
Yves DaoustYves Daoust
127k673226
$endgroup$
If $f(n)=o(1)$, by definition, for some $epsilon>0$, you will find an $N$ such that
$$n>Nimplies|f(n)|<epsilonimplies 2^{-f(n)}>2^{-epsilon}.$$
Then the tail of the sum diverges as it is bounded below by an harmonic series.
answered Dec 10 '18 at 21:55
Yves DaoustYves Daoust
127k673226
edited Dec 10 '18 at 22:08
answered Dec 10 '18 at 21:55
Yves DaoustYves Daoust
127k673226
answered Dec 10 '18 at 21:55
Yves DaoustYves Daoust
127k673226
answered Dec 10 '18 at 21:55
Yves DaoustYves Daoust
127k673226
127k673226
add a comment |
add a comment |
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$begingroup$
An $o(1)$ function cannot oscillate wildly.
$endgroup$
– Yves Daoust
Dec 10 '18 at 21:56
$begingroup$
apologies I have fixed that in the question
$endgroup$
– hologram
Dec 10 '18 at 21:59