Mixing
Proving that the series $sum_{n=1}^{infty}frac{c^{-n}}{n!}$ is convergent
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I need to show that the series $sum_{n=1}^{infty}frac{c^{-n}}{n!}$ is convergent.
I invoked the limit comparison with the series $sum_{n=1}^{infty}frac{c^n}{n!}$ which is absolutely convergent (and hence convergent).
I got $sum_{n=1}^{infty}frac{frac{c^{-n}}{n!}}{frac{c^n}{n!}}=sum_{n=1}^{infty}frac{1}{n!n!}$.
I am not sure where to go from here. Would it be correct to write that $frac{1}{n!n!}rightarrow0$ ($nrightarrowinfty$), and hence by the limit comparison test the series is convergent? Is there a better way to go about doing this?
real-analysis sequences-and-series
asked Jan 20 at 10:32
HarmanHarman
302
$endgroup$
add a comment |
$begingroup$
I need to show that the series $sum_{n=1}^{infty}frac{c^{-n}}{n!}$ is convergent.
I invoked the limit comparison with the series $sum_{n=1}^{infty}frac{c^n}{n!}$ which is absolutely convergent (and hence convergent).
I got $sum_{n=1}^{infty}frac{frac{c^{-n}}{n!}}{frac{c^n}{n!}}=sum_{n=1}^{infty}frac{1}{n!n!}$.
I am not sure where to go from here. Would it be correct to write that $frac{1}{n!n!}rightarrow0$ ($nrightarrowinfty$), and hence by the limit comparison test the series is convergent? Is there a better way to go about doing this?
real-analysis sequences-and-series
asked Jan 20 at 10:32
HarmanHarman
302
$endgroup$
2
$begingroup$
That's not how comparison test works. If you knew that $sum c^n/n!$ converges absolutely, then $sum (c^{-1})^n/n!$ converges as well.
$endgroup$
– xbh
Jan 20 at 10:38
$begingroup$
A factorial is a good sign that ratio test can be used . $ (n+1)!=n!(n+1) $
$endgroup$
– Milan Stojanovic
Jan 20 at 11:45
$begingroup$
$$frac{frac{c^{-n}}{n!}}{frac{c^n}{n!}}nefrac1{n!n!}$$
$endgroup$
– Did
Jan 31 at 13:50
add a comment |
$begingroup$
I need to show that the series $sum_{n=1}^{infty}frac{c^{-n}}{n!}$ is convergent.
I invoked the limit comparison with the series $sum_{n=1}^{infty}frac{c^n}{n!}$ which is absolutely convergent (and hence convergent).
I got $sum_{n=1}^{infty}frac{frac{c^{-n}}{n!}}{frac{c^n}{n!}}=sum_{n=1}^{infty}frac{1}{n!n!}$.
I am not sure where to go from here. Would it be correct to write that $frac{1}{n!n!}rightarrow0$ ($nrightarrowinfty$), and hence by the limit comparison test the series is convergent? Is there a better way to go about doing this?
real-analysis sequences-and-series
asked Jan 20 at 10:32
HarmanHarman
302
$endgroup$
I need to show that the series $sum_{n=1}^{infty}frac{c^{-n}}{n!}$ is convergent.
I invoked the limit comparison with the series $sum_{n=1}^{infty}frac{c^n}{n!}$ which is absolutely convergent (and hence convergent).
I got $sum_{n=1}^{infty}frac{frac{c^{-n}}{n!}}{frac{c^n}{n!}}=sum_{n=1}^{infty}frac{1}{n!n!}$.
I am not sure where to go from here. Would it be correct to write that $frac{1}{n!n!}rightarrow0$ ($nrightarrowinfty$), and hence by the limit comparison test the series is convergent? Is there a better way to go about doing this?
real-analysis sequences-and-series
real-analysis sequences-and-series
asked Jan 20 at 10:32
HarmanHarman
302
asked Jan 20 at 10:32
HarmanHarman
302
asked Jan 20 at 10:32
HarmanHarman
302
asked Jan 20 at 10:32
HarmanHarman
302
asked Jan 20 at 10:32
HarmanHarman
302
302
2
$begingroup$
That's not how comparison test works. If you knew that $sum c^n/n!$ converges absolutely, then $sum (c^{-1})^n/n!$ converges as well.
$endgroup$
– xbh
Jan 20 at 10:38
$begingroup$
A factorial is a good sign that ratio test can be used . $ (n+1)!=n!(n+1) $
$endgroup$
– Milan Stojanovic
Jan 20 at 11:45
$begingroup$
$$frac{frac{c^{-n}}{n!}}{frac{c^n}{n!}}nefrac1{n!n!}$$
$endgroup$
– Did
Jan 31 at 13:50
add a comment |
2
$begingroup$
That's not how comparison test works. If you knew that $sum c^n/n!$ converges absolutely, then $sum (c^{-1})^n/n!$ converges as well.
$endgroup$
– xbh
Jan 20 at 10:38
$begingroup$
A factorial is a good sign that ratio test can be used . $ (n+1)!=n!(n+1) $
$endgroup$
– Milan Stojanovic
Jan 20 at 11:45
$begingroup$
$$frac{frac{c^{-n}}{n!}}{frac{c^n}{n!}}nefrac1{n!n!}$$
$endgroup$
– Did
Jan 31 at 13:50
2
2
$begingroup$
That's not how comparison test works. If you knew that $sum c^n/n!$ converges absolutely, then $sum (c^{-1})^n/n!$ converges as well.
$endgroup$
– xbh
Jan 20 at 10:38
$begingroup$
That's not how comparison test works. If you knew that $sum c^n/n!$ converges absolutely, then $sum (c^{-1})^n/n!$ converges as well.
$endgroup$
– xbh
Jan 20 at 10:38
$begingroup$
A factorial is a good sign that ratio test can be used . $ (n+1)!=n!(n+1) $
$endgroup$
– Milan Stojanovic
Jan 20 at 11:45
$begingroup$
A factorial is a good sign that ratio test can be used . $ (n+1)!=n!(n+1) $
$endgroup$
– Milan Stojanovic
Jan 20 at 11:45
$begingroup$
$$frac{frac{c^{-n}}{n!}}{frac{c^n}{n!}}nefrac1{n!n!}$$
$endgroup$
– Did
Jan 31 at 13:50
$begingroup$
$$frac{frac{c^{-n}}{n!}}{frac{c^n}{n!}}nefrac1{n!n!}$$
$endgroup$
– Did
Jan 31 at 13:50
add a comment |
2 Answers
2
active
oldest
votes
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Use the ratio test:$$lim_{ninmathbb N}frac{frac{c^{-n-1}}{(n+1)!}}{frac{c^{-n}}{n!}}=lim_{ninmathbb N}frac1{ctimes(n+1)}=0<1.$$
answered Jan 20 at 10:39
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
1
$begingroup$
Provided $cne0$.
$endgroup$
– Yves Daoust
Jan 20 at 10:46
2
$begingroup$
If $c=0$, then the statement itself doesn't make sense.
$endgroup$
– José Carlos Santos
Jan 20 at 10:49
add a comment |
$begingroup$
Easier to notice that
$$ sum frac{ c^{-n} }{n!} = e^{1/c} $$
So as long as $c neq 0$, we have convergence.
answered Jan 20 at 10:35
Jimmy SabaterJimmy Sabater
2,877324
$endgroup$
1
$begingroup$
This is wrong. This series sums to $exp(1/c)$, not $exp(-1/c)$. (Note you would also need to shift the index, so the original series actually converges to $e^{1/c}-1$.
$endgroup$
– Clayton
Jan 20 at 10:59
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Use the ratio test:$$lim_{ninmathbb N}frac{frac{c^{-n-1}}{(n+1)!}}{frac{c^{-n}}{n!}}=lim_{ninmathbb N}frac1{ctimes(n+1)}=0<1.$$
answered Jan 20 at 10:39
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
1
$begingroup$
Provided $cne0$.
$endgroup$
– Yves Daoust
Jan 20 at 10:46
2
$begingroup$
If $c=0$, then the statement itself doesn't make sense.
$endgroup$
– José Carlos Santos
Jan 20 at 10:49
add a comment |
$begingroup$
Use the ratio test:$$lim_{ninmathbb N}frac{frac{c^{-n-1}}{(n+1)!}}{frac{c^{-n}}{n!}}=lim_{ninmathbb N}frac1{ctimes(n+1)}=0<1.$$
answered Jan 20 at 10:39
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
1
$begingroup$
Provided $cne0$.
$endgroup$
– Yves Daoust
Jan 20 at 10:46
2
$begingroup$
If $c=0$, then the statement itself doesn't make sense.
$endgroup$
– José Carlos Santos
Jan 20 at 10:49
add a comment |
$begingroup$
Use the ratio test:$$lim_{ninmathbb N}frac{frac{c^{-n-1}}{(n+1)!}}{frac{c^{-n}}{n!}}=lim_{ninmathbb N}frac1{ctimes(n+1)}=0<1.$$
answered Jan 20 at 10:39
José Carlos SantosJosé Carlos Santos
165k22132235
$endgroup$
Use the ratio test:$$lim_{ninmathbb N}frac{frac{c^{-n-1}}{(n+1)!}}{frac{c^{-n}}{n!}}=lim_{ninmathbb N}frac1{ctimes(n+1)}=0<1.$$
answered Jan 20 at 10:39
José Carlos SantosJosé Carlos Santos
165k22132235
answered Jan 20 at 10:39
José Carlos SantosJosé Carlos Santos
165k22132235
answered Jan 20 at 10:39
José Carlos SantosJosé Carlos Santos
165k22132235
answered Jan 20 at 10:39
José Carlos SantosJosé Carlos Santos
165k22132235
165k22132235
1
$begingroup$
Provided $cne0$.
$endgroup$
– Yves Daoust
Jan 20 at 10:46
2
$begingroup$
If $c=0$, then the statement itself doesn't make sense.
$endgroup$
– José Carlos Santos
Jan 20 at 10:49
add a comment |
1
$begingroup$
Provided $cne0$.
$endgroup$
– Yves Daoust
Jan 20 at 10:46
2
$begingroup$
If $c=0$, then the statement itself doesn't make sense.
$endgroup$
– José Carlos Santos
Jan 20 at 10:49
1
1
$begingroup$
Provided $cne0$.
$endgroup$
– Yves Daoust
Jan 20 at 10:46
$begingroup$
Provided $cne0$.
$endgroup$
– Yves Daoust
Jan 20 at 10:46
2
2
$begingroup$
If $c=0$, then the statement itself doesn't make sense.
$endgroup$
– José Carlos Santos
Jan 20 at 10:49
$begingroup$
If $c=0$, then the statement itself doesn't make sense.
$endgroup$
– José Carlos Santos
Jan 20 at 10:49
add a comment |
$begingroup$
Easier to notice that
$$ sum frac{ c^{-n} }{n!} = e^{1/c} $$
So as long as $c neq 0$, we have convergence.
answered Jan 20 at 10:35
Jimmy SabaterJimmy Sabater
2,877324
$endgroup$
1
$begingroup$
This is wrong. This series sums to $exp(1/c)$, not $exp(-1/c)$. (Note you would also need to shift the index, so the original series actually converges to $e^{1/c}-1$.
$endgroup$
– Clayton
Jan 20 at 10:59
add a comment |
$begingroup$
Easier to notice that
$$ sum frac{ c^{-n} }{n!} = e^{1/c} $$
So as long as $c neq 0$, we have convergence.
answered Jan 20 at 10:35
Jimmy SabaterJimmy Sabater
2,877324
$endgroup$
1
$begingroup$
This is wrong. This series sums to $exp(1/c)$, not $exp(-1/c)$. (Note you would also need to shift the index, so the original series actually converges to $e^{1/c}-1$.
$endgroup$
– Clayton
Jan 20 at 10:59
add a comment |
$begingroup$
Easier to notice that
$$ sum frac{ c^{-n} }{n!} = e^{1/c} $$
So as long as $c neq 0$, we have convergence.
answered Jan 20 at 10:35
Jimmy SabaterJimmy Sabater
2,877324
$endgroup$
Easier to notice that
$$ sum frac{ c^{-n} }{n!} = e^{1/c} $$
So as long as $c neq 0$, we have convergence.
answered Jan 20 at 10:35
Jimmy SabaterJimmy Sabater
2,877324
edited Jan 20 at 20:38
answered Jan 20 at 10:35
Jimmy SabaterJimmy Sabater
2,877324
answered Jan 20 at 10:35
Jimmy SabaterJimmy Sabater
2,877324
answered Jan 20 at 10:35
Jimmy SabaterJimmy Sabater
2,877324
2,877324
1
$begingroup$
This is wrong. This series sums to $exp(1/c)$, not $exp(-1/c)$. (Note you would also need to shift the index, so the original series actually converges to $e^{1/c}-1$.
$endgroup$
– Clayton
Jan 20 at 10:59
add a comment |
1
$begingroup$
This is wrong. This series sums to $exp(1/c)$, not $exp(-1/c)$. (Note you would also need to shift the index, so the original series actually converges to $e^{1/c}-1$.
$endgroup$
– Clayton
Jan 20 at 10:59
1
1
$begingroup$
This is wrong. This series sums to $exp(1/c)$, not $exp(-1/c)$. (Note you would also need to shift the index, so the original series actually converges to $e^{1/c}-1$.
$endgroup$
– Clayton
Jan 20 at 10:59
$begingroup$
This is wrong. This series sums to $exp(1/c)$, not $exp(-1/c)$. (Note you would also need to shift the index, so the original series actually converges to $e^{1/c}-1$.
$endgroup$
– Clayton
Jan 20 at 10:59
add a comment |
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2
$begingroup$
That's not how comparison test works. If you knew that $sum c^n/n!$ converges absolutely, then $sum (c^{-1})^n/n!$ converges as well.
$endgroup$
– xbh
Jan 20 at 10:38
$begingroup$
A factorial is a good sign that ratio test can be used . $ (n+1)!=n!(n+1) $
$endgroup$
– Milan Stojanovic
Jan 20 at 11:45
$begingroup$
$$frac{frac{c^{-n}}{n!}}{frac{c^n}{n!}}nefrac1{n!n!}$$
$endgroup$
– Did
Jan 31 at 13:50