calculus and series
If we suppose $lim_{nto+infty} a_n =0,$ I have to examine if $sum _{ n=1 }^{ infty } n^{frac {1}{1+a_n}}$ converges or diverges.
I used the monotone convergence theorem to show that the series diverges. If we suppose: $a_n=dfrac {1}{n},$ then $$lim_{nto+infty} a_n= lim_{nto+infty} frac {1}{n}=0. $$ Let the sequence of functions $f$ on $mathbb{N}$ be defined by $$f_n (k)= k^{frac{1}{1+frac {1}{n}}}.$$ We note that these functions are positive, and that for each $k$, as a sequence in $n, f_n (k)$ is increasing. The sum over $k$ is an integral and it's the integral over $mathbb{N}$ with respect to counting measure. In particular, by the Monotone Convergence Theorem (from the Lebesgue theory) allows to interchange limit and integral:
$$lim_{nto+infty} sum _{ k=1 }^{ infty } k^{frac {1}{1+frac {1}{n}}} = sum _{ k=1 }^{ infty } lim_{nto+infty} k^{frac {1}{1+frac {1}{n}}}= sum _{ k=1 }^{ infty } frac{1}{k}= +infty$$
and then $$ sum _{ k=1 }^{ infty } k^{frac {1}{1+frac {1}{n}}} $$ diverges. By the same argument, $$sum _{ k=1 }^{ infty } k^{frac {1}{1+a_n}}$$ diverges.
sequences-and-series
edited Jan 3 at 22:00
amWhy
192k28225439
asked Jan 3 at 21:53
George
161
|
show 1 more comment
If we suppose $lim_{nto+infty} a_n =0,$ I have to examine if $sum _{ n=1 }^{ infty } n^{frac {1}{1+a_n}}$ converges or diverges.
I used the monotone convergence theorem to show that the series diverges. If we suppose: $a_n=dfrac {1}{n},$ then $$lim_{nto+infty} a_n= lim_{nto+infty} frac {1}{n}=0. $$ Let the sequence of functions $f$ on $mathbb{N}$ be defined by $$f_n (k)= k^{frac{1}{1+frac {1}{n}}}.$$ We note that these functions are positive, and that for each $k$, as a sequence in $n, f_n (k)$ is increasing. The sum over $k$ is an integral and it's the integral over $mathbb{N}$ with respect to counting measure. In particular, by the Monotone Convergence Theorem (from the Lebesgue theory) allows to interchange limit and integral:
$$lim_{nto+infty} sum _{ k=1 }^{ infty } k^{frac {1}{1+frac {1}{n}}} = sum _{ k=1 }^{ infty } lim_{nto+infty} k^{frac {1}{1+frac {1}{n}}}= sum _{ k=1 }^{ infty } frac{1}{k}= +infty$$
and then $$ sum _{ k=1 }^{ infty } k^{frac {1}{1+frac {1}{n}}} $$ diverges. By the same argument, $$sum _{ k=1 }^{ infty } k^{frac {1}{1+a_n}}$$ diverges.
sequences-and-series
edited Jan 3 at 22:00
amWhy
192k28225439
asked Jan 3 at 21:53
George
161
OK, so you have the result for $a_n = frac{1}{n}$. What goes wrong in your proof if you take a generic $a_n$?
– user3482749
Jan 3 at 21:57
I don't know, I am wondering if I am wrong for any $$a_n$$. My professor told me that there is a more general case, but I don't know what he means
– George
Jan 3 at 22:00
George: single variables, or sequences like $a_n$, do not require two dollar-signs in mathjax. A single dollar-sign on each end is most appropriate is such context.
– amWhy
Jan 3 at 22:04
It is maybe an English expression, that I cannot understand
– George
Jan 3 at 22:05
@George Try it! Work through the proof without that assumption. Does anything go wrong? Before you get there, though, you might want to stare at your claim that $lim_{nto+infty}k^{frac{1}{1+frac{1}{n}}}=dfrac{1}{k}$.
– user3482749
Jan 3 at 22:22
|
show 1 more comment
If we suppose $lim_{nto+infty} a_n =0,$ I have to examine if $sum _{ n=1 }^{ infty } n^{frac {1}{1+a_n}}$ converges or diverges.
I used the monotone convergence theorem to show that the series diverges. If we suppose: $a_n=dfrac {1}{n},$ then $$lim_{nto+infty} a_n= lim_{nto+infty} frac {1}{n}=0. $$ Let the sequence of functions $f$ on $mathbb{N}$ be defined by $$f_n (k)= k^{frac{1}{1+frac {1}{n}}}.$$ We note that these functions are positive, and that for each $k$, as a sequence in $n, f_n (k)$ is increasing. The sum over $k$ is an integral and it's the integral over $mathbb{N}$ with respect to counting measure. In particular, by the Monotone Convergence Theorem (from the Lebesgue theory) allows to interchange limit and integral:
$$lim_{nto+infty} sum _{ k=1 }^{ infty } k^{frac {1}{1+frac {1}{n}}} = sum _{ k=1 }^{ infty } lim_{nto+infty} k^{frac {1}{1+frac {1}{n}}}= sum _{ k=1 }^{ infty } frac{1}{k}= +infty$$
and then $$ sum _{ k=1 }^{ infty } k^{frac {1}{1+frac {1}{n}}} $$ diverges. By the same argument, $$sum _{ k=1 }^{ infty } k^{frac {1}{1+a_n}}$$ diverges.
sequences-and-series
edited Jan 3 at 22:00
amWhy
192k28225439
asked Jan 3 at 21:53
George
161
If we suppose $lim_{nto+infty} a_n =0,$ I have to examine if $sum _{ n=1 }^{ infty } n^{frac {1}{1+a_n}}$ converges or diverges.
I used the monotone convergence theorem to show that the series diverges. If we suppose: $a_n=dfrac {1}{n},$ then $$lim_{nto+infty} a_n= lim_{nto+infty} frac {1}{n}=0. $$ Let the sequence of functions $f$ on $mathbb{N}$ be defined by $$f_n (k)= k^{frac{1}{1+frac {1}{n}}}.$$ We note that these functions are positive, and that for each $k$, as a sequence in $n, f_n (k)$ is increasing. The sum over $k$ is an integral and it's the integral over $mathbb{N}$ with respect to counting measure. In particular, by the Monotone Convergence Theorem (from the Lebesgue theory) allows to interchange limit and integral:
$$lim_{nto+infty} sum _{ k=1 }^{ infty } k^{frac {1}{1+frac {1}{n}}} = sum _{ k=1 }^{ infty } lim_{nto+infty} k^{frac {1}{1+frac {1}{n}}}= sum _{ k=1 }^{ infty } frac{1}{k}= +infty$$
and then $$ sum _{ k=1 }^{ infty } k^{frac {1}{1+frac {1}{n}}} $$ diverges. By the same argument, $$sum _{ k=1 }^{ infty } k^{frac {1}{1+a_n}}$$ diverges.
sequences-and-series
sequences-and-series
edited Jan 3 at 22:00
amWhy
192k28225439
asked Jan 3 at 21:53
George
161
edited Jan 3 at 22:00
amWhy
192k28225439
asked Jan 3 at 21:53
George
161
edited Jan 3 at 22:00
amWhy
192k28225439
edited Jan 3 at 22:00
amWhy
192k28225439
edited Jan 3 at 22:00
amWhy
192k28225439
192k28225439
asked Jan 3 at 21:53
George
161
asked Jan 3 at 21:53
George
161
asked Jan 3 at 21:53
George
161
161
OK, so you have the result for $a_n = frac{1}{n}$. What goes wrong in your proof if you take a generic $a_n$?
– user3482749
Jan 3 at 21:57
I don't know, I am wondering if I am wrong for any $$a_n$$. My professor told me that there is a more general case, but I don't know what he means
– George
Jan 3 at 22:00
George: single variables, or sequences like $a_n$, do not require two dollar-signs in mathjax. A single dollar-sign on each end is most appropriate is such context.
– amWhy
Jan 3 at 22:04
It is maybe an English expression, that I cannot understand
– George
Jan 3 at 22:05
@George Try it! Work through the proof without that assumption. Does anything go wrong? Before you get there, though, you might want to stare at your claim that $lim_{nto+infty}k^{frac{1}{1+frac{1}{n}}}=dfrac{1}{k}$.
– user3482749
Jan 3 at 22:22
|
show 1 more comment
OK, so you have the result for $a_n = frac{1}{n}$. What goes wrong in your proof if you take a generic $a_n$?
– user3482749
Jan 3 at 21:57
I don't know, I am wondering if I am wrong for any $$a_n$$. My professor told me that there is a more general case, but I don't know what he means
– George
Jan 3 at 22:00
George: single variables, or sequences like $a_n$, do not require two dollar-signs in mathjax. A single dollar-sign on each end is most appropriate is such context.
– amWhy
Jan 3 at 22:04
It is maybe an English expression, that I cannot understand
– George
Jan 3 at 22:05
@George Try it! Work through the proof without that assumption. Does anything go wrong? Before you get there, though, you might want to stare at your claim that $lim_{nto+infty}k^{frac{1}{1+frac{1}{n}}}=dfrac{1}{k}$.
– user3482749
Jan 3 at 22:22
OK, so you have the result for $a_n = frac{1}{n}$. What goes wrong in your proof if you take a generic $a_n$?
– user3482749
Jan 3 at 21:57
OK, so you have the result for $a_n = frac{1}{n}$. What goes wrong in your proof if you take a generic $a_n$?
– user3482749
Jan 3 at 21:57
I don't know, I am wondering if I am wrong for any $$a_n$$. My professor told me that there is a more general case, but I don't know what he means
– George
Jan 3 at 22:00
I don't know, I am wondering if I am wrong for any $$a_n$$. My professor told me that there is a more general case, but I don't know what he means
– George
Jan 3 at 22:00
George: single variables, or sequences like $a_n$, do not require two dollar-signs in mathjax. A single dollar-sign on each end is most appropriate is such context.
– amWhy
Jan 3 at 22:04
George: single variables, or sequences like $a_n$, do not require two dollar-signs in mathjax. A single dollar-sign on each end is most appropriate is such context.
– amWhy
Jan 3 at 22:04
It is maybe an English expression, that I cannot understand
– George
Jan 3 at 22:05
It is maybe an English expression, that I cannot understand
– George
Jan 3 at 22:05
@George Try it! Work through the proof without that assumption. Does anything go wrong? Before you get there, though, you might want to stare at your claim that $lim_{nto+infty}k^{frac{1}{1+frac{1}{n}}}=dfrac{1}{k}$.
– user3482749
Jan 3 at 22:22
@George Try it! Work through the proof without that assumption. Does anything go wrong? Before you get there, though, you might want to stare at your claim that $lim_{nto+infty}k^{frac{1}{1+frac{1}{n}}}=dfrac{1}{k}$.
– user3482749
Jan 3 at 22:22
|
show 1 more comment
2 Answers
2
active
oldest
votes
Hint: It is absolutely diverging because $n^x$ is equal or greater than $1$ for $n geq 1$ and $x geq 0$.
answered Jan 3 at 21:58
OmG
2,230620
add a comment |
Surely divergent. We have $$lim_{nto infty} {1over 1+a_n}=1$$ therefore for large enough $n$ we have $${1over 1+a_n}>{1over 2}$$by substitution we obtain$$n^{{1over 1+a_n}}>n^{1over 2}=sqrt n$$from which we can conclude that the series diverge since $sum sqrt n$ diverges.
answered yesterday
Mostafa Ayaz
14.1k3937
add a comment |
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2 Answers
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2 Answers
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Hint: It is absolutely diverging because $n^x$ is equal or greater than $1$ for $n geq 1$ and $x geq 0$.
answered Jan 3 at 21:58
OmG
2,230620
add a comment |
Hint: It is absolutely diverging because $n^x$ is equal or greater than $1$ for $n geq 1$ and $x geq 0$.
answered Jan 3 at 21:58
OmG
2,230620
add a comment |
Hint: It is absolutely diverging because $n^x$ is equal or greater than $1$ for $n geq 1$ and $x geq 0$.
answered Jan 3 at 21:58
OmG
2,230620
Hint: It is absolutely diverging because $n^x$ is equal or greater than $1$ for $n geq 1$ and $x geq 0$.
answered Jan 3 at 21:58
OmG
2,230620
answered Jan 3 at 21:58
OmG
2,230620
answered Jan 3 at 21:58
OmG
2,230620
answered Jan 3 at 21:58
OmG
2,230620
2,230620
add a comment |
add a comment |
Surely divergent. We have $$lim_{nto infty} {1over 1+a_n}=1$$ therefore for large enough $n$ we have $${1over 1+a_n}>{1over 2}$$by substitution we obtain$$n^{{1over 1+a_n}}>n^{1over 2}=sqrt n$$from which we can conclude that the series diverge since $sum sqrt n$ diverges.
answered yesterday
Mostafa Ayaz
14.1k3937
add a comment |
Surely divergent. We have $$lim_{nto infty} {1over 1+a_n}=1$$ therefore for large enough $n$ we have $${1over 1+a_n}>{1over 2}$$by substitution we obtain$$n^{{1over 1+a_n}}>n^{1over 2}=sqrt n$$from which we can conclude that the series diverge since $sum sqrt n$ diverges.
answered yesterday
Mostafa Ayaz
14.1k3937
add a comment |
Surely divergent. We have $$lim_{nto infty} {1over 1+a_n}=1$$ therefore for large enough $n$ we have $${1over 1+a_n}>{1over 2}$$by substitution we obtain$$n^{{1over 1+a_n}}>n^{1over 2}=sqrt n$$from which we can conclude that the series diverge since $sum sqrt n$ diverges.
answered yesterday
Mostafa Ayaz
14.1k3937
Surely divergent. We have $$lim_{nto infty} {1over 1+a_n}=1$$ therefore for large enough $n$ we have $${1over 1+a_n}>{1over 2}$$by substitution we obtain$$n^{{1over 1+a_n}}>n^{1over 2}=sqrt n$$from which we can conclude that the series diverge since $sum sqrt n$ diverges.
answered yesterday
Mostafa Ayaz
14.1k3937
answered yesterday
Mostafa Ayaz
14.1k3937
answered yesterday
Mostafa Ayaz
14.1k3937
answered yesterday
Mostafa Ayaz
14.1k3937
14.1k3937
add a comment |
add a comment |
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OK, so you have the result for $a_n = frac{1}{n}$. What goes wrong in your proof if you take a generic $a_n$?
– user3482749
Jan 3 at 21:57
I don't know, I am wondering if I am wrong for any $$a_n$$. My professor told me that there is a more general case, but I don't know what he means
– George
Jan 3 at 22:00
George: single variables, or sequences like $a_n$, do not require two dollar-signs in mathjax. A single dollar-sign on each end is most appropriate is such context.
– amWhy
Jan 3 at 22:04
It is maybe an English expression, that I cannot understand
– George
Jan 3 at 22:05
@George Try it! Work through the proof without that assumption. Does anything go wrong? Before you get there, though, you might want to stare at your claim that $lim_{nto+infty}k^{frac{1}{1+frac{1}{n}}}=dfrac{1}{k}$.
– user3482749
Jan 3 at 22:22