Why do I have to use L'Hopital in limit comparison test for $sum_{n=1}^{infty} sinleft(frac{1}{n}right)$












1














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







share|cite|improve this question






















  • 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
















1














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







share|cite|improve this question






















  • 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














1












1








1







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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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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


















  • 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










2 Answers
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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 $limlimits_{xto0}frac{sin x}x=1$?






share|cite|improve this answer





























    0














    Much less is needed.



    It suffices to know that $sin x>frac12x$ for $0<x<frac pi 2$, say.






    share|cite|improve this answer





















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      2 Answers
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      active

      oldest

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      2 Answers
      2






      active

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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 $limlimits_{xto0}frac{sin x}x=1$?






      share|cite|improve this answer


























        2














        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 $limlimits_{xto0}frac{sin x}x=1$?






        share|cite|improve this answer
























          2












          2








          2






          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 $limlimits_{xto0}frac{sin x}x=1$?






          share|cite|improve this answer












          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 $limlimits_{xto0}frac{sin x}x=1$?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 9 '18 at 21:00









          gimusi

          1




          1























              0














              Much less is needed.



              It suffices to know that $sin x>frac12x$ for $0<x<frac pi 2$, say.






              share|cite|improve this answer


























                0














                Much less is needed.



                It suffices to know that $sin x>frac12x$ for $0<x<frac pi 2$, say.






                share|cite|improve this answer
























                  0












                  0








                  0






                  Much less is needed.



                  It suffices to know that $sin x>frac12x$ for $0<x<frac pi 2$, say.






                  share|cite|improve this answer












                  Much less is needed.



                  It suffices to know that $sin x>frac12x$ for $0<x<frac pi 2$, say.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 9 '18 at 21:51









                  Hagen von Eitzen

                  276k21269496




                  276k21269496






























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