Why do I have to use L'Hopital in limit comparison test for $sum_{n=1}^{infty} sinleft(frac{1}{n}right)$
I'm trying to determine whether the following series diverges or converges:
$$sum_{n=1}^{infty} sinleft(frac{1}{n}right)$$
I use the limit comparison test, comparing it to $frac{1}{n}$, which we know diverges.
$$a_n = sinleft(frac{1}{n}right)$$
$$b_n = frac{1}{n}$$
Thus using the limit comparison test we do:
$$lim_{ntoinfty} frac{sinleft(frac{1}{n}right)}{frac{1}{n}}$$
When evaluating this limit, why can't I make it of the form:
$$lim_{ntoinfty} n sinleft(frac{1}{n}right)$$
Isn't $0 cdot infty = 0$ valid? It turns out not because the correct answer of this limit is $1$. Why is it not valid to do it the way I did it?
calculus sequences-and-series limits
asked Dec 9 '18 at 20:58
James Mitchell
26827
add a comment |
I'm trying to determine whether the following series diverges or converges:
$$sum_{n=1}^{infty} sinleft(frac{1}{n}right)$$
I use the limit comparison test, comparing it to $frac{1}{n}$, which we know diverges.
$$a_n = sinleft(frac{1}{n}right)$$
$$b_n = frac{1}{n}$$
Thus using the limit comparison test we do:
$$lim_{ntoinfty} frac{sinleft(frac{1}{n}right)}{frac{1}{n}}$$
When evaluating this limit, why can't I make it of the form:
$$lim_{ntoinfty} n sinleft(frac{1}{n}right)$$
Isn't $0 cdot infty = 0$ valid? It turns out not because the correct answer of this limit is $1$. Why is it not valid to do it the way I did it?
calculus sequences-and-series limits
asked Dec 9 '18 at 20:58
James Mitchell
26827
No, you can't write $0timesinfty=0$ because $0timesinfty$ is an indeterminate form (for example, consider $lim_{toinfty}n^a n^{-b}$ for $a,,b>0$). See also en.wikipedia.org/wiki/Indeterminate_form
– J.G.
Dec 9 '18 at 21:59
It would be a big mistake, because the limit of $nsinfrac{1}{n}$ is $1$. A form such as $inftycdot a$, with $color{red}{ane0}$, gives $infty$ or $-infty$ according to $a>0$ or $a<0$. But $inftycdot0$ cannot be treated in a simple way.
– egreg
Dec 9 '18 at 22:36
add a comment |
I'm trying to determine whether the following series diverges or converges:
$$sum_{n=1}^{infty} sinleft(frac{1}{n}right)$$
I use the limit comparison test, comparing it to $frac{1}{n}$, which we know diverges.
$$a_n = sinleft(frac{1}{n}right)$$
$$b_n = frac{1}{n}$$
Thus using the limit comparison test we do:
$$lim_{ntoinfty} frac{sinleft(frac{1}{n}right)}{frac{1}{n}}$$
When evaluating this limit, why can't I make it of the form:
$$lim_{ntoinfty} n sinleft(frac{1}{n}right)$$
Isn't $0 cdot infty = 0$ valid? It turns out not because the correct answer of this limit is $1$. Why is it not valid to do it the way I did it?
calculus sequences-and-series limits
asked Dec 9 '18 at 20:58
James Mitchell
26827
I'm trying to determine whether the following series diverges or converges:
$$sum_{n=1}^{infty} sinleft(frac{1}{n}right)$$
I use the limit comparison test, comparing it to $frac{1}{n}$, which we know diverges.
$$a_n = sinleft(frac{1}{n}right)$$
$$b_n = frac{1}{n}$$
Thus using the limit comparison test we do:
$$lim_{ntoinfty} frac{sinleft(frac{1}{n}right)}{frac{1}{n}}$$
When evaluating this limit, why can't I make it of the form:
$$lim_{ntoinfty} n sinleft(frac{1}{n}right)$$
Isn't $0 cdot infty = 0$ valid? It turns out not because the correct answer of this limit is $1$. Why is it not valid to do it the way I did it?
calculus sequences-and-series limits
calculus sequences-and-series limits
asked Dec 9 '18 at 20:58
James Mitchell
26827
asked Dec 9 '18 at 20:58
James Mitchell
26827
asked Dec 9 '18 at 20:58
James Mitchell
26827
asked Dec 9 '18 at 20:58
James Mitchell
26827
asked Dec 9 '18 at 20:58
James Mitchell
26827
26827
No, you can't write $0timesinfty=0$ because $0timesinfty$ is an indeterminate form (for example, consider $lim_{toinfty}n^a n^{-b}$ for $a,,b>0$). See also en.wikipedia.org/wiki/Indeterminate_form
– J.G.
Dec 9 '18 at 21:59
It would be a big mistake, because the limit of $nsinfrac{1}{n}$ is $1$. A form such as $inftycdot a$, with $color{red}{ane0}$, gives $infty$ or $-infty$ according to $a>0$ or $a<0$. But $inftycdot0$ cannot be treated in a simple way.
– egreg
Dec 9 '18 at 22:36
add a comment |
No, you can't write $0timesinfty=0$ because $0timesinfty$ is an indeterminate form (for example, consider $lim_{toinfty}n^a n^{-b}$ for $a,,b>0$). See also en.wikipedia.org/wiki/Indeterminate_form
– J.G.
Dec 9 '18 at 21:59
It would be a big mistake, because the limit of $nsinfrac{1}{n}$ is $1$. A form such as $inftycdot a$, with $color{red}{ane0}$, gives $infty$ or $-infty$ according to $a>0$ or $a<0$. But $inftycdot0$ cannot be treated in a simple way.
– egreg
Dec 9 '18 at 22:36
No, you can't write $0timesinfty=0$ because $0timesinfty$ is an indeterminate form (for example, consider $lim_{toinfty}n^a n^{-b}$ for $a,,b>0$). See also en.wikipedia.org/wiki/Indeterminate_form
– J.G.
Dec 9 '18 at 21:59
No, you can't write $0timesinfty=0$ because $0timesinfty$ is an indeterminate form (for example, consider $lim_{toinfty}n^a n^{-b}$ for $a,,b>0$). See also en.wikipedia.org/wiki/Indeterminate_form
– J.G.
Dec 9 '18 at 21:59
It would be a big mistake, because the limit of $nsinfrac{1}{n}$ is $1$. A form such as $inftycdot a$, with $color{red}{ane0}$, gives $infty$ or $-infty$ according to $a>0$ or $a<0$. But $inftycdot0$ cannot be treated in a simple way.
– egreg
Dec 9 '18 at 22:36
It would be a big mistake, because the limit of $nsinfrac{1}{n}$ is $1$. A form such as $inftycdot a$, with $color{red}{ane0}$, gives $infty$ or $-infty$ according to $a>0$ or $a<0$. But $inftycdot0$ cannot be treated in a simple way.
– egreg
Dec 9 '18 at 22:36
add a comment |
2 Answers
2
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Indeed we don't need to use l'Hopital at all, simply take $x=frac 1 n to 0$ to obtain
$$frac{a_n}{b_n}=frac{ sinleft(frac{1}{n}right)}{frac1n}=frac{sin x}{x}$$
and refer to standard limits.
You shouldn't have doubts about that if you are dealing now with series, in case refer to the related
- How to prove that $limlimits_{xto0}frac{sin x}x=1$?
answered Dec 9 '18 at 21:00
gimusi
1
add a comment |
Much less is needed.
It suffices to know that $sin x>frac12x$ for $0<x<frac pi 2$, say.
answered Dec 9 '18 at 21:51
Hagen von Eitzen
276k21269496
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Indeed we don't need to use l'Hopital at all, simply take $x=frac 1 n to 0$ to obtain
$$frac{a_n}{b_n}=frac{ sinleft(frac{1}{n}right)}{frac1n}=frac{sin x}{x}$$
and refer to standard limits.
You shouldn't have doubts about that if you are dealing now with series, in case refer to the related
- How to prove that $limlimits_{xto0}frac{sin x}x=1$?
answered Dec 9 '18 at 21:00
gimusi
1
add a comment |
Indeed we don't need to use l'Hopital at all, simply take $x=frac 1 n to 0$ to obtain
$$frac{a_n}{b_n}=frac{ sinleft(frac{1}{n}right)}{frac1n}=frac{sin x}{x}$$
and refer to standard limits.
You shouldn't have doubts about that if you are dealing now with series, in case refer to the related
- How to prove that $limlimits_{xto0}frac{sin x}x=1$?
answered Dec 9 '18 at 21:00
gimusi
1
add a comment |
Indeed we don't need to use l'Hopital at all, simply take $x=frac 1 n to 0$ to obtain
$$frac{a_n}{b_n}=frac{ sinleft(frac{1}{n}right)}{frac1n}=frac{sin x}{x}$$
and refer to standard limits.
You shouldn't have doubts about that if you are dealing now with series, in case refer to the related
- How to prove that $limlimits_{xto0}frac{sin x}x=1$?
answered Dec 9 '18 at 21:00
gimusi
1
Indeed we don't need to use l'Hopital at all, simply take $x=frac 1 n to 0$ to obtain
$$frac{a_n}{b_n}=frac{ sinleft(frac{1}{n}right)}{frac1n}=frac{sin x}{x}$$
and refer to standard limits.
You shouldn't have doubts about that if you are dealing now with series, in case refer to the related
- How to prove that $limlimits_{xto0}frac{sin x}x=1$?
answered Dec 9 '18 at 21:00
gimusi
1
answered Dec 9 '18 at 21:00
gimusi
1
answered Dec 9 '18 at 21:00
gimusi
1
answered Dec 9 '18 at 21:00
gimusi
1
1
add a comment |
add a comment |
Much less is needed.
It suffices to know that $sin x>frac12x$ for $0<x<frac pi 2$, say.
answered Dec 9 '18 at 21:51
Hagen von Eitzen
276k21269496
add a comment |
Much less is needed.
It suffices to know that $sin x>frac12x$ for $0<x<frac pi 2$, say.
answered Dec 9 '18 at 21:51
Hagen von Eitzen
276k21269496
add a comment |
Much less is needed.
It suffices to know that $sin x>frac12x$ for $0<x<frac pi 2$, say.
answered Dec 9 '18 at 21:51
Hagen von Eitzen
276k21269496
Much less is needed.
It suffices to know that $sin x>frac12x$ for $0<x<frac pi 2$, say.
answered Dec 9 '18 at 21:51
Hagen von Eitzen
276k21269496
answered Dec 9 '18 at 21:51
Hagen von Eitzen
276k21269496
answered Dec 9 '18 at 21:51
Hagen von Eitzen
276k21269496
answered Dec 9 '18 at 21:51
Hagen von Eitzen
276k21269496
276k21269496
add a comment |
add a comment |
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No, you can't write $0timesinfty=0$ because $0timesinfty$ is an indeterminate form (for example, consider $lim_{toinfty}n^a n^{-b}$ for $a,,b>0$). See also en.wikipedia.org/wiki/Indeterminate_form
– J.G.
Dec 9 '18 at 21:59
It would be a big mistake, because the limit of $nsinfrac{1}{n}$ is $1$. A form such as $inftycdot a$, with $color{red}{ane0}$, gives $infty$ or $-infty$ according to $a>0$ or $a<0$. But $inftycdot0$ cannot be treated in a simple way.
– egreg
Dec 9 '18 at 22:36