Mixing
Using ratiotest to check convergence of $sum sin(frac{1}{n!})$.
$begingroup$
I'm preparing a class on the convergence of series and how to check it using de ratiotest. One of the exercises asks to determine the convergence of the series
$$sum sin(frac{1}{n!})$$
using the ratiotest.
I need to compute the following limit:
$$lim_{n to infty} frac{sin(frac{1}{(n+1)!})}{sin(frac{1}{n!})}$$
but I have no idea how to do this: I tried using that $sin(x) leq x$, the sandwich theorem,...
I also can't recall any usefull geometric identity to split of a factor $sin(frac{1}{n!})$.
Any help would be very appreciated.
sequences-and-series trigonometry convergence
edited Jan 22 at 19:37
José Carlos Santos
166k22132235
asked Jan 22 at 15:47
StudentStudent
2,1441727
$endgroup$
add a comment |
$begingroup$
I'm preparing a class on the convergence of series and how to check it using de ratiotest. One of the exercises asks to determine the convergence of the series
$$sum sin(frac{1}{n!})$$
using the ratiotest.
I need to compute the following limit:
$$lim_{n to infty} frac{sin(frac{1}{(n+1)!})}{sin(frac{1}{n!})}$$
but I have no idea how to do this: I tried using that $sin(x) leq x$, the sandwich theorem,...
I also can't recall any usefull geometric identity to split of a factor $sin(frac{1}{n!})$.
Any help would be very appreciated.
sequences-and-series trigonometry convergence
edited Jan 22 at 19:37
José Carlos Santos
166k22132235
asked Jan 22 at 15:47
StudentStudent
2,1441727
$endgroup$
$begingroup$
When $0 leq x leq pi/2$, $2x/pi leq sin{x} leq x$.
$endgroup$
– Mindlack
Jan 22 at 15:49
$begingroup$
@Mindlack Thank you.
$endgroup$
– Student
Jan 22 at 15:57
add a comment |
$begingroup$
I'm preparing a class on the convergence of series and how to check it using de ratiotest. One of the exercises asks to determine the convergence of the series
$$sum sin(frac{1}{n!})$$
using the ratiotest.
I need to compute the following limit:
$$lim_{n to infty} frac{sin(frac{1}{(n+1)!})}{sin(frac{1}{n!})}$$
but I have no idea how to do this: I tried using that $sin(x) leq x$, the sandwich theorem,...
I also can't recall any usefull geometric identity to split of a factor $sin(frac{1}{n!})$.
Any help would be very appreciated.
sequences-and-series trigonometry convergence
edited Jan 22 at 19:37
José Carlos Santos
166k22132235
asked Jan 22 at 15:47
StudentStudent
2,1441727
$endgroup$
I'm preparing a class on the convergence of series and how to check it using de ratiotest. One of the exercises asks to determine the convergence of the series
$$sum sin(frac{1}{n!})$$
using the ratiotest.
I need to compute the following limit:
$$lim_{n to infty} frac{sin(frac{1}{(n+1)!})}{sin(frac{1}{n!})}$$
but I have no idea how to do this: I tried using that $sin(x) leq x$, the sandwich theorem,...
I also can't recall any usefull geometric identity to split of a factor $sin(frac{1}{n!})$.
Any help would be very appreciated.
sequences-and-series trigonometry convergence
sequences-and-series trigonometry convergence
edited Jan 22 at 19:37
José Carlos Santos
166k22132235
asked Jan 22 at 15:47
StudentStudent
2,1441727
edited Jan 22 at 19:37
José Carlos Santos
166k22132235
asked Jan 22 at 15:47
StudentStudent
2,1441727
edited Jan 22 at 19:37
José Carlos Santos
166k22132235
edited Jan 22 at 19:37
José Carlos Santos
166k22132235
edited Jan 22 at 19:37
José Carlos Santos
166k22132235
166k22132235
asked Jan 22 at 15:47
StudentStudent
2,1441727
asked Jan 22 at 15:47
StudentStudent
2,1441727
asked Jan 22 at 15:47
StudentStudent
2,1441727
2,1441727
$begingroup$
When $0 leq x leq pi/2$, $2x/pi leq sin{x} leq x$.
$endgroup$
– Mindlack
Jan 22 at 15:49
$begingroup$
@Mindlack Thank you.
$endgroup$
– Student
Jan 22 at 15:57
add a comment |
$begingroup$
When $0 leq x leq pi/2$, $2x/pi leq sin{x} leq x$.
$endgroup$
– Mindlack
Jan 22 at 15:49
$begingroup$
@Mindlack Thank you.
$endgroup$
– Student
Jan 22 at 15:57
$begingroup$
When $0 leq x leq pi/2$, $2x/pi leq sin{x} leq x$.
$endgroup$
– Mindlack
Jan 22 at 15:49
$begingroup$
When $0 leq x leq pi/2$, $2x/pi leq sin{x} leq x$.
$endgroup$
– Mindlack
Jan 22 at 15:49
$begingroup$
@Mindlack Thank you.
$endgroup$
– Student
Jan 22 at 15:57
$begingroup$
@Mindlack Thank you.
$endgroup$
– Student
Jan 22 at 15:57
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
$$dfrac{sinleft(dfrac{1}{(n+1)!}right)}{sinleft(dfrac{1}{n!}right)}=dfrac{sinleft(dfrac{1}{(n+1)!}right)}{dfrac{1}{(n+1)!}}timesdfrac{dfrac{1}{n!}}{sinleft(dfrac{1}{n!}right)}timesdfrac{1}{n+1}$$
answered Jan 22 at 15:53
ScientificaScientifica
6,82941335
$endgroup$
$begingroup$
Thanks a lot! Can't believe I forgot about that.
$endgroup$
– Student
Jan 22 at 15:56
$begingroup$
@Student It's a pleasure :)
$endgroup$
– Scientifica
Jan 22 at 16:04
add a comment |
$begingroup$
Note that in this particular case, since $sin(frac 1{n!})>0$ you have directly from $sin(x)le x$
$$0lesumlimits_{n=0}^{infty}sin(frac 1{n!})lesumlimits_{n=0}^{infty}frac 1{n!}=e$$
So the series is absolutely convergent (since $nearrow$ and bounded above). But ok, if it is an exercise on the ratio test, so be it.
answered Jan 22 at 20:05
zwimzwim
12.5k831
$endgroup$
$begingroup$
Thanks for your answer. In a previous class, these students have actually learned that increasing bounded sequences have a limit, so I could give this to them as some extra explanation, taking for granted that the last equality holds :)
$endgroup$
– Student
Jan 22 at 21:06
add a comment |
$begingroup$
Note thatbegin{align}frac{sinleft(frac1{(n+1)!}right)}{sinleft(frac1{n!}right)}&=frac{frac{sinleft(frac1{(n+1)!}right)}{frac1{(n+1)!}}}{frac{sinleft(frac1{n!}right)}{frac1{(n+1)!}}}\&=frac1{n+1}timesfrac{frac{sinleft(frac1{(n+1)!}right)}{frac1{(n+1)!}}}{frac{sinleft(frac1{n!}right)}{frac1{n!}}}.end{align}Since$$lim_{ntoinfty}frac{frac{sinleft(frac1{(n+1)!}right)}{frac1{(n+1)!}}}{frac{sinleft(frac1{n!}right)}{frac1{n!}}}=frac11=1,$$you know that$$lim_{ntoinfty}frac1{n+1}timesfrac{frac{sinleft(frac1{(n+1)!}right)}{frac1{(n+1)!}}}{frac{sinleft(frac1{n!}right)}{frac1{n!}}}=0.$$
answered Jan 22 at 15:53
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
$begingroup$
Thanks a lot, can't believe I forgot about that.
$endgroup$
– Student
Jan 22 at 15:56
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$dfrac{sinleft(dfrac{1}{(n+1)!}right)}{sinleft(dfrac{1}{n!}right)}=dfrac{sinleft(dfrac{1}{(n+1)!}right)}{dfrac{1}{(n+1)!}}timesdfrac{dfrac{1}{n!}}{sinleft(dfrac{1}{n!}right)}timesdfrac{1}{n+1}$$
answered Jan 22 at 15:53
ScientificaScientifica
6,82941335
$endgroup$
$begingroup$
Thanks a lot! Can't believe I forgot about that.
$endgroup$
– Student
Jan 22 at 15:56
$begingroup$
@Student It's a pleasure :)
$endgroup$
– Scientifica
Jan 22 at 16:04
add a comment |
$begingroup$
$$dfrac{sinleft(dfrac{1}{(n+1)!}right)}{sinleft(dfrac{1}{n!}right)}=dfrac{sinleft(dfrac{1}{(n+1)!}right)}{dfrac{1}{(n+1)!}}timesdfrac{dfrac{1}{n!}}{sinleft(dfrac{1}{n!}right)}timesdfrac{1}{n+1}$$
answered Jan 22 at 15:53
ScientificaScientifica
6,82941335
$endgroup$
$begingroup$
Thanks a lot! Can't believe I forgot about that.
$endgroup$
– Student
Jan 22 at 15:56
$begingroup$
@Student It's a pleasure :)
$endgroup$
– Scientifica
Jan 22 at 16:04
add a comment |
$begingroup$
$$dfrac{sinleft(dfrac{1}{(n+1)!}right)}{sinleft(dfrac{1}{n!}right)}=dfrac{sinleft(dfrac{1}{(n+1)!}right)}{dfrac{1}{(n+1)!}}timesdfrac{dfrac{1}{n!}}{sinleft(dfrac{1}{n!}right)}timesdfrac{1}{n+1}$$
answered Jan 22 at 15:53
ScientificaScientifica
6,82941335
$endgroup$
$$dfrac{sinleft(dfrac{1}{(n+1)!}right)}{sinleft(dfrac{1}{n!}right)}=dfrac{sinleft(dfrac{1}{(n+1)!}right)}{dfrac{1}{(n+1)!}}timesdfrac{dfrac{1}{n!}}{sinleft(dfrac{1}{n!}right)}timesdfrac{1}{n+1}$$
answered Jan 22 at 15:53
ScientificaScientifica
6,82941335
answered Jan 22 at 15:53
ScientificaScientifica
6,82941335
answered Jan 22 at 15:53
ScientificaScientifica
6,82941335
answered Jan 22 at 15:53
ScientificaScientifica
6,82941335
6,82941335
$begingroup$
Thanks a lot! Can't believe I forgot about that.
$endgroup$
– Student
Jan 22 at 15:56
$begingroup$
@Student It's a pleasure :)
$endgroup$
– Scientifica
Jan 22 at 16:04
add a comment |
$begingroup$
Thanks a lot! Can't believe I forgot about that.
$endgroup$
– Student
Jan 22 at 15:56
$begingroup$
@Student It's a pleasure :)
$endgroup$
– Scientifica
Jan 22 at 16:04
$begingroup$
Thanks a lot! Can't believe I forgot about that.
$endgroup$
– Student
Jan 22 at 15:56
$begingroup$
Thanks a lot! Can't believe I forgot about that.
$endgroup$
– Student
Jan 22 at 15:56
$begingroup$
@Student It's a pleasure :)
$endgroup$
– Scientifica
Jan 22 at 16:04
$begingroup$
@Student It's a pleasure :)
$endgroup$
– Scientifica
Jan 22 at 16:04
add a comment |
$begingroup$
Note that in this particular case, since $sin(frac 1{n!})>0$ you have directly from $sin(x)le x$
$$0lesumlimits_{n=0}^{infty}sin(frac 1{n!})lesumlimits_{n=0}^{infty}frac 1{n!}=e$$
So the series is absolutely convergent (since $nearrow$ and bounded above). But ok, if it is an exercise on the ratio test, so be it.
answered Jan 22 at 20:05
zwimzwim
12.5k831
$endgroup$
$begingroup$
Thanks for your answer. In a previous class, these students have actually learned that increasing bounded sequences have a limit, so I could give this to them as some extra explanation, taking for granted that the last equality holds :)
$endgroup$
– Student
Jan 22 at 21:06
add a comment |
$begingroup$
Note that in this particular case, since $sin(frac 1{n!})>0$ you have directly from $sin(x)le x$
$$0lesumlimits_{n=0}^{infty}sin(frac 1{n!})lesumlimits_{n=0}^{infty}frac 1{n!}=e$$
So the series is absolutely convergent (since $nearrow$ and bounded above). But ok, if it is an exercise on the ratio test, so be it.
answered Jan 22 at 20:05
zwimzwim
12.5k831
$endgroup$
$begingroup$
Thanks for your answer. In a previous class, these students have actually learned that increasing bounded sequences have a limit, so I could give this to them as some extra explanation, taking for granted that the last equality holds :)
$endgroup$
– Student
Jan 22 at 21:06
add a comment |
$begingroup$
Note that in this particular case, since $sin(frac 1{n!})>0$ you have directly from $sin(x)le x$
$$0lesumlimits_{n=0}^{infty}sin(frac 1{n!})lesumlimits_{n=0}^{infty}frac 1{n!}=e$$
So the series is absolutely convergent (since $nearrow$ and bounded above). But ok, if it is an exercise on the ratio test, so be it.
answered Jan 22 at 20:05
zwimzwim
12.5k831
$endgroup$
Note that in this particular case, since $sin(frac 1{n!})>0$ you have directly from $sin(x)le x$
$$0lesumlimits_{n=0}^{infty}sin(frac 1{n!})lesumlimits_{n=0}^{infty}frac 1{n!}=e$$
So the series is absolutely convergent (since $nearrow$ and bounded above). But ok, if it is an exercise on the ratio test, so be it.
answered Jan 22 at 20:05
zwimzwim
12.5k831
answered Jan 22 at 20:05
zwimzwim
12.5k831
answered Jan 22 at 20:05
zwimzwim
12.5k831
answered Jan 22 at 20:05
zwimzwim
12.5k831
12.5k831
$begingroup$
Thanks for your answer. In a previous class, these students have actually learned that increasing bounded sequences have a limit, so I could give this to them as some extra explanation, taking for granted that the last equality holds :)
$endgroup$
– Student
Jan 22 at 21:06
add a comment |
$begingroup$
Thanks for your answer. In a previous class, these students have actually learned that increasing bounded sequences have a limit, so I could give this to them as some extra explanation, taking for granted that the last equality holds :)
$endgroup$
– Student
Jan 22 at 21:06
$begingroup$
Thanks for your answer. In a previous class, these students have actually learned that increasing bounded sequences have a limit, so I could give this to them as some extra explanation, taking for granted that the last equality holds :)
$endgroup$
– Student
Jan 22 at 21:06
$begingroup$
Thanks for your answer. In a previous class, these students have actually learned that increasing bounded sequences have a limit, so I could give this to them as some extra explanation, taking for granted that the last equality holds :)
$endgroup$
– Student
Jan 22 at 21:06
add a comment |
$begingroup$
Note thatbegin{align}frac{sinleft(frac1{(n+1)!}right)}{sinleft(frac1{n!}right)}&=frac{frac{sinleft(frac1{(n+1)!}right)}{frac1{(n+1)!}}}{frac{sinleft(frac1{n!}right)}{frac1{(n+1)!}}}\&=frac1{n+1}timesfrac{frac{sinleft(frac1{(n+1)!}right)}{frac1{(n+1)!}}}{frac{sinleft(frac1{n!}right)}{frac1{n!}}}.end{align}Since$$lim_{ntoinfty}frac{frac{sinleft(frac1{(n+1)!}right)}{frac1{(n+1)!}}}{frac{sinleft(frac1{n!}right)}{frac1{n!}}}=frac11=1,$$you know that$$lim_{ntoinfty}frac1{n+1}timesfrac{frac{sinleft(frac1{(n+1)!}right)}{frac1{(n+1)!}}}{frac{sinleft(frac1{n!}right)}{frac1{n!}}}=0.$$
answered Jan 22 at 15:53
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
$begingroup$
Thanks a lot, can't believe I forgot about that.
$endgroup$
– Student
Jan 22 at 15:56
add a comment |
$begingroup$
Note thatbegin{align}frac{sinleft(frac1{(n+1)!}right)}{sinleft(frac1{n!}right)}&=frac{frac{sinleft(frac1{(n+1)!}right)}{frac1{(n+1)!}}}{frac{sinleft(frac1{n!}right)}{frac1{(n+1)!}}}\&=frac1{n+1}timesfrac{frac{sinleft(frac1{(n+1)!}right)}{frac1{(n+1)!}}}{frac{sinleft(frac1{n!}right)}{frac1{n!}}}.end{align}Since$$lim_{ntoinfty}frac{frac{sinleft(frac1{(n+1)!}right)}{frac1{(n+1)!}}}{frac{sinleft(frac1{n!}right)}{frac1{n!}}}=frac11=1,$$you know that$$lim_{ntoinfty}frac1{n+1}timesfrac{frac{sinleft(frac1{(n+1)!}right)}{frac1{(n+1)!}}}{frac{sinleft(frac1{n!}right)}{frac1{n!}}}=0.$$
answered Jan 22 at 15:53
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
$begingroup$
Thanks a lot, can't believe I forgot about that.
$endgroup$
– Student
Jan 22 at 15:56
add a comment |
$begingroup$
Note thatbegin{align}frac{sinleft(frac1{(n+1)!}right)}{sinleft(frac1{n!}right)}&=frac{frac{sinleft(frac1{(n+1)!}right)}{frac1{(n+1)!}}}{frac{sinleft(frac1{n!}right)}{frac1{(n+1)!}}}\&=frac1{n+1}timesfrac{frac{sinleft(frac1{(n+1)!}right)}{frac1{(n+1)!}}}{frac{sinleft(frac1{n!}right)}{frac1{n!}}}.end{align}Since$$lim_{ntoinfty}frac{frac{sinleft(frac1{(n+1)!}right)}{frac1{(n+1)!}}}{frac{sinleft(frac1{n!}right)}{frac1{n!}}}=frac11=1,$$you know that$$lim_{ntoinfty}frac1{n+1}timesfrac{frac{sinleft(frac1{(n+1)!}right)}{frac1{(n+1)!}}}{frac{sinleft(frac1{n!}right)}{frac1{n!}}}=0.$$
answered Jan 22 at 15:53
José Carlos SantosJosé Carlos Santos
166k22132235
$endgroup$
Note thatbegin{align}frac{sinleft(frac1{(n+1)!}right)}{sinleft(frac1{n!}right)}&=frac{frac{sinleft(frac1{(n+1)!}right)}{frac1{(n+1)!}}}{frac{sinleft(frac1{n!}right)}{frac1{(n+1)!}}}\&=frac1{n+1}timesfrac{frac{sinleft(frac1{(n+1)!}right)}{frac1{(n+1)!}}}{frac{sinleft(frac1{n!}right)}{frac1{n!}}}.end{align}Since$$lim_{ntoinfty}frac{frac{sinleft(frac1{(n+1)!}right)}{frac1{(n+1)!}}}{frac{sinleft(frac1{n!}right)}{frac1{n!}}}=frac11=1,$$you know that$$lim_{ntoinfty}frac1{n+1}timesfrac{frac{sinleft(frac1{(n+1)!}right)}{frac1{(n+1)!}}}{frac{sinleft(frac1{n!}right)}{frac1{n!}}}=0.$$
answered Jan 22 at 15:53
José Carlos SantosJosé Carlos Santos
166k22132235
answered Jan 22 at 15:53
José Carlos SantosJosé Carlos Santos
166k22132235
answered Jan 22 at 15:53
José Carlos SantosJosé Carlos Santos
166k22132235
answered Jan 22 at 15:53
José Carlos SantosJosé Carlos Santos
166k22132235
166k22132235
$begingroup$
Thanks a lot, can't believe I forgot about that.
$endgroup$
– Student
Jan 22 at 15:56
add a comment |
$begingroup$
Thanks a lot, can't believe I forgot about that.
$endgroup$
– Student
Jan 22 at 15:56
$begingroup$
Thanks a lot, can't believe I forgot about that.
$endgroup$
– Student
Jan 22 at 15:56
$begingroup$
Thanks a lot, can't believe I forgot about that.
$endgroup$
– Student
Jan 22 at 15:56
add a comment |
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$begingroup$
When $0 leq x leq pi/2$, $2x/pi leq sin{x} leq x$.
$endgroup$
– Mindlack
Jan 22 at 15:49
$begingroup$
@Mindlack Thank you.
$endgroup$
– Student
Jan 22 at 15:57