How to prove $lim_{ntoinfty} sum_{k=n+1}^{infty} frac{r^{k}}{k!} = 0$? [closed]
How to prove $$lim_{ntoinfty} sum_{k=n+1}^{infty} frac{r^{k}}{k!} = 0
?$$
I found that it is very similar to the Taylor series formula for $e^{r}$. Is there a nice way to show that this equals $0$?
sequences-and-series limits
asked Dec 3 '18 at 18:53
josephjoseph
4329
closed as off-topic by user21820, user10354138, ancientmathematician, Brahadeesh, Rebellos Dec 12 '18 at 9:34
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, user10354138, ancientmathematician, Brahadeesh, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
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How to prove $$lim_{ntoinfty} sum_{k=n+1}^{infty} frac{r^{k}}{k!} = 0
?$$
I found that it is very similar to the Taylor series formula for $e^{r}$. Is there a nice way to show that this equals $0$?
sequences-and-series limits
asked Dec 3 '18 at 18:53
josephjoseph
4329
closed as off-topic by user21820, user10354138, ancientmathematician, Brahadeesh, Rebellos Dec 12 '18 at 9:34
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, user10354138, ancientmathematician, Brahadeesh, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
4
Since $ sum_{n=0}^infty frac{r^k}{k!} $ converges (to $e^r$), its tail must go to zero.
– MisterRiemann
Dec 3 '18 at 18:55
why is that true?
– joseph
Dec 3 '18 at 18:56
2
@joseph to get some intuition, think about what happens when you sum a constant series -- over infinite amount of terms...
– gt6989b
Dec 3 '18 at 18:57
6
$$ e^r - sum_{k=0}^n frac{r^k}{k!} = sum_{k=n+1}^infty frac{r^k}{k!}. $$ Now let $ntoinfty$ on both sides.
– MisterRiemann
Dec 3 '18 at 18:58
1
@MisterRiemann I think it's a very elegant answer to a problem that others might encounter as well
– Andrei
Dec 3 '18 at 19:22
|
show 4 more comments
How to prove $$lim_{ntoinfty} sum_{k=n+1}^{infty} frac{r^{k}}{k!} = 0
?$$
I found that it is very similar to the Taylor series formula for $e^{r}$. Is there a nice way to show that this equals $0$?
sequences-and-series limits
asked Dec 3 '18 at 18:53
josephjoseph
4329
How to prove $$lim_{ntoinfty} sum_{k=n+1}^{infty} frac{r^{k}}{k!} = 0
?$$
I found that it is very similar to the Taylor series formula for $e^{r}$. Is there a nice way to show that this equals $0$?
sequences-and-series limits
sequences-and-series limits
asked Dec 3 '18 at 18:53
josephjoseph
4329
asked Dec 3 '18 at 18:53
josephjoseph
4329
asked Dec 3 '18 at 18:53
josephjoseph
4329
asked Dec 3 '18 at 18:53
josephjoseph
4329
asked Dec 3 '18 at 18:53
josephjoseph
4329
4329
closed as off-topic by user21820, user10354138, ancientmathematician, Brahadeesh, Rebellos Dec 12 '18 at 9:34
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, user10354138, ancientmathematician, Brahadeesh, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by user21820, user10354138, ancientmathematician, Brahadeesh, Rebellos Dec 12 '18 at 9:34
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, user10354138, ancientmathematician, Brahadeesh, Rebellos
If this question can be reworded to fit the rules in the help center, please edit the question.
4
Since $ sum_{n=0}^infty frac{r^k}{k!} $ converges (to $e^r$), its tail must go to zero.
– MisterRiemann
Dec 3 '18 at 18:55
why is that true?
– joseph
Dec 3 '18 at 18:56
2
@joseph to get some intuition, think about what happens when you sum a constant series -- over infinite amount of terms...
– gt6989b
Dec 3 '18 at 18:57
6
$$ e^r - sum_{k=0}^n frac{r^k}{k!} = sum_{k=n+1}^infty frac{r^k}{k!}. $$ Now let $ntoinfty$ on both sides.
– MisterRiemann
Dec 3 '18 at 18:58
1
@MisterRiemann I think it's a very elegant answer to a problem that others might encounter as well
– Andrei
Dec 3 '18 at 19:22
|
show 4 more comments
4
Since $ sum_{n=0}^infty frac{r^k}{k!} $ converges (to $e^r$), its tail must go to zero.
– MisterRiemann
Dec 3 '18 at 18:55
why is that true?
– joseph
Dec 3 '18 at 18:56
2
@joseph to get some intuition, think about what happens when you sum a constant series -- over infinite amount of terms...
– gt6989b
Dec 3 '18 at 18:57
6
$$ e^r - sum_{k=0}^n frac{r^k}{k!} = sum_{k=n+1}^infty frac{r^k}{k!}. $$ Now let $ntoinfty$ on both sides.
– MisterRiemann
Dec 3 '18 at 18:58
1
@MisterRiemann I think it's a very elegant answer to a problem that others might encounter as well
– Andrei
Dec 3 '18 at 19:22
4
4
Since $ sum_{n=0}^infty frac{r^k}{k!} $ converges (to $e^r$), its tail must go to zero.
– MisterRiemann
Dec 3 '18 at 18:55
Since $ sum_{n=0}^infty frac{r^k}{k!} $ converges (to $e^r$), its tail must go to zero.
– MisterRiemann
Dec 3 '18 at 18:55
why is that true?
– joseph
Dec 3 '18 at 18:56
why is that true?
– joseph
Dec 3 '18 at 18:56
2
2
@joseph to get some intuition, think about what happens when you sum a constant series -- over infinite amount of terms...
– gt6989b
Dec 3 '18 at 18:57
@joseph to get some intuition, think about what happens when you sum a constant series -- over infinite amount of terms...
– gt6989b
Dec 3 '18 at 18:57
6
6
$$ e^r - sum_{k=0}^n frac{r^k}{k!} = sum_{k=n+1}^infty frac{r^k}{k!}. $$ Now let $ntoinfty$ on both sides.
– MisterRiemann
Dec 3 '18 at 18:58
$$ e^r - sum_{k=0}^n frac{r^k}{k!} = sum_{k=n+1}^infty frac{r^k}{k!}. $$ Now let $ntoinfty$ on both sides.
– MisterRiemann
Dec 3 '18 at 18:58
1
1
@MisterRiemann I think it's a very elegant answer to a problem that others might encounter as well
– Andrei
Dec 3 '18 at 19:22
@MisterRiemann I think it's a very elegant answer to a problem that others might encounter as well
– Andrei
Dec 3 '18 at 19:22
|
show 4 more comments
3 Answers
3
active
oldest
votes
Since $$ sum_{k=0}^infty frac{r^k}{k!} = e^r, $$
the tail of the sum on the LHS must go to zero. To see this, note that
$$ sum_{k=n+1}^infty frac{r^k}{k!} = e^r - sum_{k=0}^n frac{r^k}{k!}, $$
and let $n to infty$ on both sides.
answered Dec 3 '18 at 19:24
MisterRiemannMisterRiemann
5,8841624
4
This argument is potentially circular; if you define $e^r$ in terms of this series then you need to know that it converges for the definition to make sense, which is equivalent to the remainder going to zero, which is precisely the question.
– Qiaochu Yuan
Dec 3 '18 at 22:05
1
@QiaochuYuan Fair point.
– MisterRiemann
Dec 3 '18 at 22:24
I will be more than glad to remove this answer if the author of the question can unaccept it. I must say that I agree with the comment above, and I feel that the answer by @RRL is more suitable for this problem.
– MisterRiemann
Dec 5 '18 at 20:33
add a comment |
Saying $S - S_n to 0$ because $S_n to S$ is not untrue, but it is somewhat vacuous and certainly not a first-principles argument for this question.
Note that $lim_{k to infty} r^k/(k-2)! = 0$ (factorial dominates exponential).
For sufficiently large $n$ we have $r^k/(k-2)! < 1/2$ and
$$sum_{k=n+1}^infty frac{r^k}{k!} < frac{1}{2}sum_{k=n+1}^infty frac{1}{k(k-1)} = frac{1}{2n} to_{n to infty} 0$$
answered Dec 3 '18 at 19:14
RRLRRL
49.3k42573
add a comment |
Once $k$ exceeds $r$, take some value of $k>r$; call it $kappa$. The terms thence diminish faster than they would if you were simply multiplying by ${r overkappa}$, as the value of $k$ is incremented at each new term. And the convergence of the sum of a decreasing geometric progression is elementary.
answered Dec 3 '18 at 19:10
AmbretteOrriseyAmbretteOrrisey
57410
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Since $$ sum_{k=0}^infty frac{r^k}{k!} = e^r, $$
the tail of the sum on the LHS must go to zero. To see this, note that
$$ sum_{k=n+1}^infty frac{r^k}{k!} = e^r - sum_{k=0}^n frac{r^k}{k!}, $$
and let $n to infty$ on both sides.
answered Dec 3 '18 at 19:24
MisterRiemannMisterRiemann
5,8841624
4
This argument is potentially circular; if you define $e^r$ in terms of this series then you need to know that it converges for the definition to make sense, which is equivalent to the remainder going to zero, which is precisely the question.
– Qiaochu Yuan
Dec 3 '18 at 22:05
1
@QiaochuYuan Fair point.
– MisterRiemann
Dec 3 '18 at 22:24
I will be more than glad to remove this answer if the author of the question can unaccept it. I must say that I agree with the comment above, and I feel that the answer by @RRL is more suitable for this problem.
– MisterRiemann
Dec 5 '18 at 20:33
add a comment |
Since $$ sum_{k=0}^infty frac{r^k}{k!} = e^r, $$
the tail of the sum on the LHS must go to zero. To see this, note that
$$ sum_{k=n+1}^infty frac{r^k}{k!} = e^r - sum_{k=0}^n frac{r^k}{k!}, $$
and let $n to infty$ on both sides.
answered Dec 3 '18 at 19:24
MisterRiemannMisterRiemann
5,8841624
4
This argument is potentially circular; if you define $e^r$ in terms of this series then you need to know that it converges for the definition to make sense, which is equivalent to the remainder going to zero, which is precisely the question.
– Qiaochu Yuan
Dec 3 '18 at 22:05
1
@QiaochuYuan Fair point.
– MisterRiemann
Dec 3 '18 at 22:24
I will be more than glad to remove this answer if the author of the question can unaccept it. I must say that I agree with the comment above, and I feel that the answer by @RRL is more suitable for this problem.
– MisterRiemann
Dec 5 '18 at 20:33
add a comment |
Since $$ sum_{k=0}^infty frac{r^k}{k!} = e^r, $$
the tail of the sum on the LHS must go to zero. To see this, note that
$$ sum_{k=n+1}^infty frac{r^k}{k!} = e^r - sum_{k=0}^n frac{r^k}{k!}, $$
and let $n to infty$ on both sides.
answered Dec 3 '18 at 19:24
MisterRiemannMisterRiemann
5,8841624
Since $$ sum_{k=0}^infty frac{r^k}{k!} = e^r, $$
the tail of the sum on the LHS must go to zero. To see this, note that
$$ sum_{k=n+1}^infty frac{r^k}{k!} = e^r - sum_{k=0}^n frac{r^k}{k!}, $$
and let $n to infty$ on both sides.
answered Dec 3 '18 at 19:24
MisterRiemannMisterRiemann
5,8841624
answered Dec 3 '18 at 19:24
MisterRiemannMisterRiemann
5,8841624
answered Dec 3 '18 at 19:24
MisterRiemannMisterRiemann
5,8841624
answered Dec 3 '18 at 19:24
MisterRiemannMisterRiemann
5,8841624
5,8841624
4
This argument is potentially circular; if you define $e^r$ in terms of this series then you need to know that it converges for the definition to make sense, which is equivalent to the remainder going to zero, which is precisely the question.
– Qiaochu Yuan
Dec 3 '18 at 22:05
1
@QiaochuYuan Fair point.
– MisterRiemann
Dec 3 '18 at 22:24
I will be more than glad to remove this answer if the author of the question can unaccept it. I must say that I agree with the comment above, and I feel that the answer by @RRL is more suitable for this problem.
– MisterRiemann
Dec 5 '18 at 20:33
add a comment |
4
This argument is potentially circular; if you define $e^r$ in terms of this series then you need to know that it converges for the definition to make sense, which is equivalent to the remainder going to zero, which is precisely the question.
– Qiaochu Yuan
Dec 3 '18 at 22:05
1
@QiaochuYuan Fair point.
– MisterRiemann
Dec 3 '18 at 22:24
I will be more than glad to remove this answer if the author of the question can unaccept it. I must say that I agree with the comment above, and I feel that the answer by @RRL is more suitable for this problem.
– MisterRiemann
Dec 5 '18 at 20:33
4
4
This argument is potentially circular; if you define $e^r$ in terms of this series then you need to know that it converges for the definition to make sense, which is equivalent to the remainder going to zero, which is precisely the question.
– Qiaochu Yuan
Dec 3 '18 at 22:05
This argument is potentially circular; if you define $e^r$ in terms of this series then you need to know that it converges for the definition to make sense, which is equivalent to the remainder going to zero, which is precisely the question.
– Qiaochu Yuan
Dec 3 '18 at 22:05
1
1
@QiaochuYuan Fair point.
– MisterRiemann
Dec 3 '18 at 22:24
@QiaochuYuan Fair point.
– MisterRiemann
Dec 3 '18 at 22:24
I will be more than glad to remove this answer if the author of the question can unaccept it. I must say that I agree with the comment above, and I feel that the answer by @RRL is more suitable for this problem.
– MisterRiemann
Dec 5 '18 at 20:33
I will be more than glad to remove this answer if the author of the question can unaccept it. I must say that I agree with the comment above, and I feel that the answer by @RRL is more suitable for this problem.
– MisterRiemann
Dec 5 '18 at 20:33
add a comment |
Saying $S - S_n to 0$ because $S_n to S$ is not untrue, but it is somewhat vacuous and certainly not a first-principles argument for this question.
Note that $lim_{k to infty} r^k/(k-2)! = 0$ (factorial dominates exponential).
For sufficiently large $n$ we have $r^k/(k-2)! < 1/2$ and
$$sum_{k=n+1}^infty frac{r^k}{k!} < frac{1}{2}sum_{k=n+1}^infty frac{1}{k(k-1)} = frac{1}{2n} to_{n to infty} 0$$
answered Dec 3 '18 at 19:14
RRLRRL
49.3k42573
add a comment |
Saying $S - S_n to 0$ because $S_n to S$ is not untrue, but it is somewhat vacuous and certainly not a first-principles argument for this question.
Note that $lim_{k to infty} r^k/(k-2)! = 0$ (factorial dominates exponential).
For sufficiently large $n$ we have $r^k/(k-2)! < 1/2$ and
$$sum_{k=n+1}^infty frac{r^k}{k!} < frac{1}{2}sum_{k=n+1}^infty frac{1}{k(k-1)} = frac{1}{2n} to_{n to infty} 0$$
answered Dec 3 '18 at 19:14
RRLRRL
49.3k42573
add a comment |
Saying $S - S_n to 0$ because $S_n to S$ is not untrue, but it is somewhat vacuous and certainly not a first-principles argument for this question.
Note that $lim_{k to infty} r^k/(k-2)! = 0$ (factorial dominates exponential).
For sufficiently large $n$ we have $r^k/(k-2)! < 1/2$ and
$$sum_{k=n+1}^infty frac{r^k}{k!} < frac{1}{2}sum_{k=n+1}^infty frac{1}{k(k-1)} = frac{1}{2n} to_{n to infty} 0$$
answered Dec 3 '18 at 19:14
RRLRRL
49.3k42573
Saying $S - S_n to 0$ because $S_n to S$ is not untrue, but it is somewhat vacuous and certainly not a first-principles argument for this question.
Note that $lim_{k to infty} r^k/(k-2)! = 0$ (factorial dominates exponential).
For sufficiently large $n$ we have $r^k/(k-2)! < 1/2$ and
$$sum_{k=n+1}^infty frac{r^k}{k!} < frac{1}{2}sum_{k=n+1}^infty frac{1}{k(k-1)} = frac{1}{2n} to_{n to infty} 0$$
answered Dec 3 '18 at 19:14
RRLRRL
49.3k42573
edited Dec 3 '18 at 22:17
answered Dec 3 '18 at 19:14
RRLRRL
49.3k42573
answered Dec 3 '18 at 19:14
RRLRRL
49.3k42573
answered Dec 3 '18 at 19:14
RRLRRL
49.3k42573
49.3k42573
add a comment |
add a comment |
Once $k$ exceeds $r$, take some value of $k>r$; call it $kappa$. The terms thence diminish faster than they would if you were simply multiplying by ${r overkappa}$, as the value of $k$ is incremented at each new term. And the convergence of the sum of a decreasing geometric progression is elementary.
answered Dec 3 '18 at 19:10
AmbretteOrriseyAmbretteOrrisey
57410
add a comment |
Once $k$ exceeds $r$, take some value of $k>r$; call it $kappa$. The terms thence diminish faster than they would if you were simply multiplying by ${r overkappa}$, as the value of $k$ is incremented at each new term. And the convergence of the sum of a decreasing geometric progression is elementary.
answered Dec 3 '18 at 19:10
AmbretteOrriseyAmbretteOrrisey
57410
add a comment |
Once $k$ exceeds $r$, take some value of $k>r$; call it $kappa$. The terms thence diminish faster than they would if you were simply multiplying by ${r overkappa}$, as the value of $k$ is incremented at each new term. And the convergence of the sum of a decreasing geometric progression is elementary.
answered Dec 3 '18 at 19:10
AmbretteOrriseyAmbretteOrrisey
57410
Once $k$ exceeds $r$, take some value of $k>r$; call it $kappa$. The terms thence diminish faster than they would if you were simply multiplying by ${r overkappa}$, as the value of $k$ is incremented at each new term. And the convergence of the sum of a decreasing geometric progression is elementary.
answered Dec 3 '18 at 19:10
AmbretteOrriseyAmbretteOrrisey
57410
answered Dec 3 '18 at 19:10
AmbretteOrriseyAmbretteOrrisey
57410
answered Dec 3 '18 at 19:10
AmbretteOrriseyAmbretteOrrisey
57410
answered Dec 3 '18 at 19:10
AmbretteOrriseyAmbretteOrrisey
57410
57410
add a comment |
add a comment |
4
Since $ sum_{n=0}^infty frac{r^k}{k!} $ converges (to $e^r$), its tail must go to zero.
– MisterRiemann
Dec 3 '18 at 18:55
why is that true?
– joseph
Dec 3 '18 at 18:56
2
@joseph to get some intuition, think about what happens when you sum a constant series -- over infinite amount of terms...
– gt6989b
Dec 3 '18 at 18:57
6
$$ e^r - sum_{k=0}^n frac{r^k}{k!} = sum_{k=n+1}^infty frac{r^k}{k!}. $$ Now let $ntoinfty$ on both sides.
– MisterRiemann
Dec 3 '18 at 18:58
1
@MisterRiemann I think it's a very elegant answer to a problem that others might encounter as well
– Andrei
Dec 3 '18 at 19:22