Mixing
Show that $lim_{nrightarrowinfty}e^{-n}=0$
$begingroup$
I wrote out the term from above and get
$text{Show that: }lim_{nrightarrowinfty}sum_{k=0}^{infty}(-1)^kfrac{n^k}{k!}=0$
I can use Leibniz' Criterion to find out that it converges but I don't know nothing about the value. Is there a $0$ convergent upper bound for this expression. Can somebody give me a hint how to find it please?
Edit: I have changed the fraction in the expression
sequences-and-series limits
edited Jan 23 at 14:42
Bernard
122k741116
asked Jan 23 at 14:25
RM777RM777
38312
$endgroup$
add a comment |
$begingroup$
I wrote out the term from above and get
$text{Show that: }lim_{nrightarrowinfty}sum_{k=0}^{infty}(-1)^kfrac{n^k}{k!}=0$
I can use Leibniz' Criterion to find out that it converges but I don't know nothing about the value. Is there a $0$ convergent upper bound for this expression. Can somebody give me a hint how to find it please?
Edit: I have changed the fraction in the expression
sequences-and-series limits
edited Jan 23 at 14:42
Bernard
122k741116
asked Jan 23 at 14:25
RM777RM777
38312
$endgroup$
1
$begingroup$
$e^{-n} = frac{1}{e^n}$ also $2<e$ and $2^nrightarrowinfty$ so $frac{1}{2^n}rightarrow 0$ and $1/e<1/2$
$endgroup$
– Yanko
Jan 23 at 14:27
4
$begingroup$
This is just a special case of the general fact: if $|x|<1$ then $lim_{n to infty} x^n=0$.
$endgroup$
– Lee Mosher
Jan 23 at 14:39
1
$begingroup$
Although $lim_{nrightarrowinfty}sum_{k=0}^{infty}(-1)^kfrac{n^k}{k!}=0 $ is true, you cannot prove it by looking at the terms of this series.
$endgroup$
– GEdgar
Jan 23 at 15:04
add a comment |
$begingroup$
I wrote out the term from above and get
$text{Show that: }lim_{nrightarrowinfty}sum_{k=0}^{infty}(-1)^kfrac{n^k}{k!}=0$
I can use Leibniz' Criterion to find out that it converges but I don't know nothing about the value. Is there a $0$ convergent upper bound for this expression. Can somebody give me a hint how to find it please?
Edit: I have changed the fraction in the expression
sequences-and-series limits
edited Jan 23 at 14:42
Bernard
122k741116
asked Jan 23 at 14:25
RM777RM777
38312
$endgroup$
I wrote out the term from above and get
$text{Show that: }lim_{nrightarrowinfty}sum_{k=0}^{infty}(-1)^kfrac{n^k}{k!}=0$
I can use Leibniz' Criterion to find out that it converges but I don't know nothing about the value. Is there a $0$ convergent upper bound for this expression. Can somebody give me a hint how to find it please?
Edit: I have changed the fraction in the expression
sequences-and-series limits
sequences-and-series limits
edited Jan 23 at 14:42
Bernard
122k741116
asked Jan 23 at 14:25
RM777RM777
38312
edited Jan 23 at 14:42
Bernard
122k741116
asked Jan 23 at 14:25
RM777RM777
38312
edited Jan 23 at 14:42
Bernard
122k741116
edited Jan 23 at 14:42
Bernard
122k741116
edited Jan 23 at 14:42
Bernard
122k741116
122k741116
asked Jan 23 at 14:25
RM777RM777
38312
asked Jan 23 at 14:25
RM777RM777
38312
asked Jan 23 at 14:25
RM777RM777
38312
38312
1
$begingroup$
$e^{-n} = frac{1}{e^n}$ also $2<e$ and $2^nrightarrowinfty$ so $frac{1}{2^n}rightarrow 0$ and $1/e<1/2$
$endgroup$
– Yanko
Jan 23 at 14:27
4
$begingroup$
This is just a special case of the general fact: if $|x|<1$ then $lim_{n to infty} x^n=0$.
$endgroup$
– Lee Mosher
Jan 23 at 14:39
1
$begingroup$
Although $lim_{nrightarrowinfty}sum_{k=0}^{infty}(-1)^kfrac{n^k}{k!}=0 $ is true, you cannot prove it by looking at the terms of this series.
$endgroup$
– GEdgar
Jan 23 at 15:04
add a comment |
1
$begingroup$
$e^{-n} = frac{1}{e^n}$ also $2<e$ and $2^nrightarrowinfty$ so $frac{1}{2^n}rightarrow 0$ and $1/e<1/2$
$endgroup$
– Yanko
Jan 23 at 14:27
4
$begingroup$
This is just a special case of the general fact: if $|x|<1$ then $lim_{n to infty} x^n=0$.
$endgroup$
– Lee Mosher
Jan 23 at 14:39
1
$begingroup$
Although $lim_{nrightarrowinfty}sum_{k=0}^{infty}(-1)^kfrac{n^k}{k!}=0 $ is true, you cannot prove it by looking at the terms of this series.
$endgroup$
– GEdgar
Jan 23 at 15:04
1
1
$begingroup$
$e^{-n} = frac{1}{e^n}$ also $2<e$ and $2^nrightarrowinfty$ so $frac{1}{2^n}rightarrow 0$ and $1/e<1/2$
$endgroup$
– Yanko
Jan 23 at 14:27
$begingroup$
$e^{-n} = frac{1}{e^n}$ also $2<e$ and $2^nrightarrowinfty$ so $frac{1}{2^n}rightarrow 0$ and $1/e<1/2$
$endgroup$
– Yanko
Jan 23 at 14:27
4
4
$begingroup$
This is just a special case of the general fact: if $|x|<1$ then $lim_{n to infty} x^n=0$.
$endgroup$
– Lee Mosher
Jan 23 at 14:39
$begingroup$
This is just a special case of the general fact: if $|x|<1$ then $lim_{n to infty} x^n=0$.
$endgroup$
– Lee Mosher
Jan 23 at 14:39
1
1
$begingroup$
Although $lim_{nrightarrowinfty}sum_{k=0}^{infty}(-1)^kfrac{n^k}{k!}=0 $ is true, you cannot prove it by looking at the terms of this series.
$endgroup$
– GEdgar
Jan 23 at 15:04
$begingroup$
Although $lim_{nrightarrowinfty}sum_{k=0}^{infty}(-1)^kfrac{n^k}{k!}=0 $ is true, you cannot prove it by looking at the terms of this series.
$endgroup$
– GEdgar
Jan 23 at 15:04
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Here is the standard elementary proof that
$a^n to 0$
if $0 < a < 1$.
$a =1/(1+b)$ where $b = 1/a -1 > 0$.
By Bernoulli's inequality,
$(1+b)^n ge 1+bn > bn$
so
$a^n =1/(1+b)^n
< 1/(bn)
to 0$
as $n to infty$.
answered Jan 23 at 14:53
marty cohenmarty cohen
74.3k549128
$endgroup$
$begingroup$
Okay I understood it now however Prior to that I had to understand that the Approximation of $e$ is between 2 and 3- Like @GEdgar said in the comment it is apperantly not very easy to proof the Statement by looking at the Definition of the exponential function, ie. the Terms of the series.math.stackexchange.com/questions/2987534/… math.stackexchange.com/questions/3084932/…
$endgroup$
– RM777
Jan 27 at 13:51
add a comment |
$begingroup$
For any $a$ such that $0 < a < 1$, you can find natural $n$ to make $a^n$ smaller than any positive number $x$:
$a^n < x iff n , log(a) < log(x) iff n > log(x)/log(a)$
Also, $a^n > 0$ for any natural $n$.
These 2 facts imply $lim_{n -> infty} , a^n = 0$
answered Jan 23 at 14:46
JaviJavi
3979
$endgroup$
add a comment |
$begingroup$
Do not expand, just use
$$e^{-n}=(e^{-1})^n$$ which is quickly decreasing.
answered Jan 23 at 14:57
Yves DaoustYves Daoust
130k676227
$endgroup$
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
$begingroup$
Here is the standard elementary proof that
$a^n to 0$
if $0 < a < 1$.
$a =1/(1+b)$ where $b = 1/a -1 > 0$.
By Bernoulli's inequality,
$(1+b)^n ge 1+bn > bn$
so
$a^n =1/(1+b)^n
< 1/(bn)
to 0$
as $n to infty$.
answered Jan 23 at 14:53
marty cohenmarty cohen
74.3k549128
$endgroup$
$begingroup$
Okay I understood it now however Prior to that I had to understand that the Approximation of $e$ is between 2 and 3- Like @GEdgar said in the comment it is apperantly not very easy to proof the Statement by looking at the Definition of the exponential function, ie. the Terms of the series.math.stackexchange.com/questions/2987534/… math.stackexchange.com/questions/3084932/…
$endgroup$
– RM777
Jan 27 at 13:51
add a comment |
$begingroup$
Here is the standard elementary proof that
$a^n to 0$
if $0 < a < 1$.
$a =1/(1+b)$ where $b = 1/a -1 > 0$.
By Bernoulli's inequality,
$(1+b)^n ge 1+bn > bn$
so
$a^n =1/(1+b)^n
< 1/(bn)
to 0$
as $n to infty$.
answered Jan 23 at 14:53
marty cohenmarty cohen
74.3k549128
$endgroup$
$begingroup$
Okay I understood it now however Prior to that I had to understand that the Approximation of $e$ is between 2 and 3- Like @GEdgar said in the comment it is apperantly not very easy to proof the Statement by looking at the Definition of the exponential function, ie. the Terms of the series.math.stackexchange.com/questions/2987534/… math.stackexchange.com/questions/3084932/…
$endgroup$
– RM777
Jan 27 at 13:51
add a comment |
$begingroup$
Here is the standard elementary proof that
$a^n to 0$
if $0 < a < 1$.
$a =1/(1+b)$ where $b = 1/a -1 > 0$.
By Bernoulli's inequality,
$(1+b)^n ge 1+bn > bn$
so
$a^n =1/(1+b)^n
< 1/(bn)
to 0$
as $n to infty$.
answered Jan 23 at 14:53
marty cohenmarty cohen
74.3k549128
$endgroup$
Here is the standard elementary proof that
$a^n to 0$
if $0 < a < 1$.
$a =1/(1+b)$ where $b = 1/a -1 > 0$.
By Bernoulli's inequality,
$(1+b)^n ge 1+bn > bn$
so
$a^n =1/(1+b)^n
< 1/(bn)
to 0$
as $n to infty$.
answered Jan 23 at 14:53
marty cohenmarty cohen
74.3k549128
answered Jan 23 at 14:53
marty cohenmarty cohen
74.3k549128
answered Jan 23 at 14:53
marty cohenmarty cohen
74.3k549128
answered Jan 23 at 14:53
marty cohenmarty cohen
74.3k549128
74.3k549128
$begingroup$
Okay I understood it now however Prior to that I had to understand that the Approximation of $e$ is between 2 and 3- Like @GEdgar said in the comment it is apperantly not very easy to proof the Statement by looking at the Definition of the exponential function, ie. the Terms of the series.math.stackexchange.com/questions/2987534/… math.stackexchange.com/questions/3084932/…
$endgroup$
– RM777
Jan 27 at 13:51
add a comment |
$begingroup$
Okay I understood it now however Prior to that I had to understand that the Approximation of $e$ is between 2 and 3- Like @GEdgar said in the comment it is apperantly not very easy to proof the Statement by looking at the Definition of the exponential function, ie. the Terms of the series.math.stackexchange.com/questions/2987534/… math.stackexchange.com/questions/3084932/…
$endgroup$
– RM777
Jan 27 at 13:51
$begingroup$
Okay I understood it now however Prior to that I had to understand that the Approximation of $e$ is between 2 and 3- Like @GEdgar said in the comment it is apperantly not very easy to proof the Statement by looking at the Definition of the exponential function, ie. the Terms of the series.math.stackexchange.com/questions/2987534/… math.stackexchange.com/questions/3084932/…
$endgroup$
– RM777
Jan 27 at 13:51
$begingroup$
Okay I understood it now however Prior to that I had to understand that the Approximation of $e$ is between 2 and 3- Like @GEdgar said in the comment it is apperantly not very easy to proof the Statement by looking at the Definition of the exponential function, ie. the Terms of the series.math.stackexchange.com/questions/2987534/… math.stackexchange.com/questions/3084932/…
$endgroup$
– RM777
Jan 27 at 13:51
add a comment |
$begingroup$
For any $a$ such that $0 < a < 1$, you can find natural $n$ to make $a^n$ smaller than any positive number $x$:
$a^n < x iff n , log(a) < log(x) iff n > log(x)/log(a)$
Also, $a^n > 0$ for any natural $n$.
These 2 facts imply $lim_{n -> infty} , a^n = 0$
answered Jan 23 at 14:46
JaviJavi
3979
$endgroup$
add a comment |
$begingroup$
For any $a$ such that $0 < a < 1$, you can find natural $n$ to make $a^n$ smaller than any positive number $x$:
$a^n < x iff n , log(a) < log(x) iff n > log(x)/log(a)$
Also, $a^n > 0$ for any natural $n$.
These 2 facts imply $lim_{n -> infty} , a^n = 0$
answered Jan 23 at 14:46
JaviJavi
3979
$endgroup$
add a comment |
$begingroup$
For any $a$ such that $0 < a < 1$, you can find natural $n$ to make $a^n$ smaller than any positive number $x$:
$a^n < x iff n , log(a) < log(x) iff n > log(x)/log(a)$
Also, $a^n > 0$ for any natural $n$.
These 2 facts imply $lim_{n -> infty} , a^n = 0$
answered Jan 23 at 14:46
JaviJavi
3979
$endgroup$
For any $a$ such that $0 < a < 1$, you can find natural $n$ to make $a^n$ smaller than any positive number $x$:
$a^n < x iff n , log(a) < log(x) iff n > log(x)/log(a)$
Also, $a^n > 0$ for any natural $n$.
These 2 facts imply $lim_{n -> infty} , a^n = 0$
answered Jan 23 at 14:46
JaviJavi
3979
answered Jan 23 at 14:46
JaviJavi
3979
answered Jan 23 at 14:46
JaviJavi
3979
answered Jan 23 at 14:46
JaviJavi
3979
3979
add a comment |
add a comment |
$begingroup$
Do not expand, just use
$$e^{-n}=(e^{-1})^n$$ which is quickly decreasing.
answered Jan 23 at 14:57
Yves DaoustYves Daoust
130k676227
$endgroup$
add a comment |
$begingroup$
Do not expand, just use
$$e^{-n}=(e^{-1})^n$$ which is quickly decreasing.
answered Jan 23 at 14:57
Yves DaoustYves Daoust
130k676227
$endgroup$
add a comment |
$begingroup$
Do not expand, just use
$$e^{-n}=(e^{-1})^n$$ which is quickly decreasing.
answered Jan 23 at 14:57
Yves DaoustYves Daoust
130k676227
$endgroup$
Do not expand, just use
$$e^{-n}=(e^{-1})^n$$ which is quickly decreasing.
answered Jan 23 at 14:57
Yves DaoustYves Daoust
130k676227
answered Jan 23 at 14:57
Yves DaoustYves Daoust
130k676227
answered Jan 23 at 14:57
Yves DaoustYves Daoust
130k676227
answered Jan 23 at 14:57
Yves DaoustYves Daoust
130k676227
130k676227
add a comment |
add a comment |
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1
$begingroup$
$e^{-n} = frac{1}{e^n}$ also $2<e$ and $2^nrightarrowinfty$ so $frac{1}{2^n}rightarrow 0$ and $1/e<1/2$
$endgroup$
– Yanko
Jan 23 at 14:27
4
$begingroup$
This is just a special case of the general fact: if $|x|<1$ then $lim_{n to infty} x^n=0$.
$endgroup$
– Lee Mosher
Jan 23 at 14:39
1
$begingroup$
Although $lim_{nrightarrowinfty}sum_{k=0}^{infty}(-1)^kfrac{n^k}{k!}=0 $ is true, you cannot prove it by looking at the terms of this series.
$endgroup$
– GEdgar
Jan 23 at 15:04