Find the sum of the series $sumfrac1(n+1)(n+2)(n+3)…(n+k)$
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This question was in my maths paper . I tried to prove that the series converges but failed to do so. Any hint are welcomed .
sequences-and-series
asked Sep 9 at 10:24
user580093
8815
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This question was in my maths paper . I tried to prove that the series converges but failed to do so. Any hint are welcomed .
sequences-and-series
asked Sep 9 at 10:24
user580093
8815
2
Is the sum made over $n$ or over $k$?
– LucaMac
Sep 9 at 10:26
@LucaMac over $k$
– user580093
Sep 9 at 10:27
So it's $sumlimits_k=1^+infty frac1(n+1)cdotldotscdot(n+k)$, right?
– LucaMac
Sep 9 at 10:28
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
This question was in my maths paper . I tried to prove that the series converges but failed to do so. Any hint are welcomed .
sequences-and-series
asked Sep 9 at 10:24
user580093
8815
This question was in my maths paper . I tried to prove that the series converges but failed to do so. Any hint are welcomed .
sequences-and-series
sequences-and-series
asked Sep 9 at 10:24
user580093
8815
asked Sep 9 at 10:24
user580093
8815
asked Sep 9 at 10:24
user580093
8815
asked Sep 9 at 10:24
user580093
8815
asked Sep 9 at 10:24
user580093
8815
8815
2
Is the sum made over $n$ or over $k$?
– LucaMac
Sep 9 at 10:26
@LucaMac over $k$
– user580093
Sep 9 at 10:27
So it's $sumlimits_k=1^+infty frac1(n+1)cdotldotscdot(n+k)$, right?
– LucaMac
Sep 9 at 10:28
add a comment |Â
2
Is the sum made over $n$ or over $k$?
– LucaMac
Sep 9 at 10:26
@LucaMac over $k$
– user580093
Sep 9 at 10:27
So it's $sumlimits_k=1^+infty frac1(n+1)cdotldotscdot(n+k)$, right?
– LucaMac
Sep 9 at 10:28
2
2
Is the sum made over $n$ or over $k$?
– LucaMac
Sep 9 at 10:26
Is the sum made over $n$ or over $k$?
– LucaMac
Sep 9 at 10:26
@LucaMac over $k$
– user580093
Sep 9 at 10:27
@LucaMac over $k$
– user580093
Sep 9 at 10:27
So it's $sumlimits_k=1^+infty frac1(n+1)cdotldotscdot(n+k)$, right?
– LucaMac
Sep 9 at 10:28
So it's $sumlimits_k=1^+infty frac1(n+1)cdotldotscdot(n+k)$, right?
– LucaMac
Sep 9 at 10:28
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
$$sum_k = 1^inftyfrac1(n+1)(n+2)(n+3)…(n+k) = n!sum_k = 1^inftyfrac1(n+k)!$$
Which clearly converges super-exponentially. We can express this in terms of generalized Gamma function is desired.
answered Sep 9 at 10:44
Brevan Ellefsen
11.6k31549
add a comment |Â
up vote
0
down vote
Hint: $$(n+1)cdot(n+2)cdots(n+k)=frac(n+k)!n!$$
from here you only need to get to $displaystyle sum frac1k!=e$, which is easy by adding some missing terms.
answered Sep 9 at 10:43
b00n heT
8,90211833
for $displaystyle sum frac1k!=e$ to be true, $k$ needs to start from $0$, otherwise $displaystyle sumlimits_k=1 frac1k!=e-1$
– rtybase
Sep 9 at 11:58
1
Indeed. I nowhere specified where the summation starts, though, I just suggested the idea
– b00n heT
Sep 9 at 12:02
1
The reason I raised it was, for $n=0$ it should lead to $sumlimits_k=1 frac1k!$ and from my calculations it should be $leq 2$, but then $eleq 2$ doesn't makes sense :) ... it took me ~10 minutes to realise it's $e-1<2$ in fact.
– rtybase
Sep 9 at 12:07
add a comment |Â
up vote
0
down vote
Just focusing on convergence, since $n+kgeq n+2, forall kgeq2,forall ngeq0$ we have:
$$0<frac1(n+1)(n+2)(n+3)...(n+k)leqfrac1(n+1)(n+2)^k-1$$
thus
$$sumlimits_k=1frac1(n+1)(n+2)(n+3)...(n+k)leq
frac1n+1+sumlimits_k=2frac1(n+1)(n+2)^k-1=\
frac1n+1left(1+sumlimits_colorredk=2frac1(n+2)^k-1right)=
frac1n+1left(1+sumlimits_colorredk=1frac1(n+2)^kright)=\
frac1n+1left(sumlimits_colorredk=0frac1(n+2)^kright)=...$$
the latter is an infinite geometric progression, or:
$$...=frac1n+1cdot frac11-frac1n+2=fracn+2(n+1)^2$$
answered Sep 9 at 11:57
rtybase
9,21721433
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$$sum_k = 1^inftyfrac1(n+1)(n+2)(n+3)…(n+k) = n!sum_k = 1^inftyfrac1(n+k)!$$
Which clearly converges super-exponentially. We can express this in terms of generalized Gamma function is desired.
answered Sep 9 at 10:44
Brevan Ellefsen
11.6k31549
add a comment |Â
up vote
1
down vote
accepted
$$sum_k = 1^inftyfrac1(n+1)(n+2)(n+3)…(n+k) = n!sum_k = 1^inftyfrac1(n+k)!$$
Which clearly converges super-exponentially. We can express this in terms of generalized Gamma function is desired.
answered Sep 9 at 10:44
Brevan Ellefsen
11.6k31549
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$$sum_k = 1^inftyfrac1(n+1)(n+2)(n+3)…(n+k) = n!sum_k = 1^inftyfrac1(n+k)!$$
Which clearly converges super-exponentially. We can express this in terms of generalized Gamma function is desired.
answered Sep 9 at 10:44
Brevan Ellefsen
11.6k31549
$$sum_k = 1^inftyfrac1(n+1)(n+2)(n+3)…(n+k) = n!sum_k = 1^inftyfrac1(n+k)!$$
Which clearly converges super-exponentially. We can express this in terms of generalized Gamma function is desired.
answered Sep 9 at 10:44
Brevan Ellefsen
11.6k31549
answered Sep 9 at 10:44
Brevan Ellefsen
11.6k31549
answered Sep 9 at 10:44
Brevan Ellefsen
11.6k31549
answered Sep 9 at 10:44
Brevan Ellefsen
11.6k31549
11.6k31549
add a comment |Â
add a comment |Â
up vote
0
down vote
Hint: $$(n+1)cdot(n+2)cdots(n+k)=frac(n+k)!n!$$
from here you only need to get to $displaystyle sum frac1k!=e$, which is easy by adding some missing terms.
answered Sep 9 at 10:43
b00n heT
8,90211833
for $displaystyle sum frac1k!=e$ to be true, $k$ needs to start from $0$, otherwise $displaystyle sumlimits_k=1 frac1k!=e-1$
– rtybase
Sep 9 at 11:58
1
Indeed. I nowhere specified where the summation starts, though, I just suggested the idea
– b00n heT
Sep 9 at 12:02
1
The reason I raised it was, for $n=0$ it should lead to $sumlimits_k=1 frac1k!$ and from my calculations it should be $leq 2$, but then $eleq 2$ doesn't makes sense :) ... it took me ~10 minutes to realise it's $e-1<2$ in fact.
– rtybase
Sep 9 at 12:07
add a comment |Â
up vote
0
down vote
Hint: $$(n+1)cdot(n+2)cdots(n+k)=frac(n+k)!n!$$
from here you only need to get to $displaystyle sum frac1k!=e$, which is easy by adding some missing terms.
answered Sep 9 at 10:43
b00n heT
8,90211833
for $displaystyle sum frac1k!=e$ to be true, $k$ needs to start from $0$, otherwise $displaystyle sumlimits_k=1 frac1k!=e-1$
– rtybase
Sep 9 at 11:58
1
Indeed. I nowhere specified where the summation starts, though, I just suggested the idea
– b00n heT
Sep 9 at 12:02
1
The reason I raised it was, for $n=0$ it should lead to $sumlimits_k=1 frac1k!$ and from my calculations it should be $leq 2$, but then $eleq 2$ doesn't makes sense :) ... it took me ~10 minutes to realise it's $e-1<2$ in fact.
– rtybase
Sep 9 at 12:07
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint: $$(n+1)cdot(n+2)cdots(n+k)=frac(n+k)!n!$$
from here you only need to get to $displaystyle sum frac1k!=e$, which is easy by adding some missing terms.
answered Sep 9 at 10:43
b00n heT
8,90211833
Hint: $$(n+1)cdot(n+2)cdots(n+k)=frac(n+k)!n!$$
from here you only need to get to $displaystyle sum frac1k!=e$, which is easy by adding some missing terms.
answered Sep 9 at 10:43
b00n heT
8,90211833
answered Sep 9 at 10:43
b00n heT
8,90211833
answered Sep 9 at 10:43
b00n heT
8,90211833
answered Sep 9 at 10:43
b00n heT
8,90211833
8,90211833
for $displaystyle sum frac1k!=e$ to be true, $k$ needs to start from $0$, otherwise $displaystyle sumlimits_k=1 frac1k!=e-1$
– rtybase
Sep 9 at 11:58
1
Indeed. I nowhere specified where the summation starts, though, I just suggested the idea
– b00n heT
Sep 9 at 12:02
1
The reason I raised it was, for $n=0$ it should lead to $sumlimits_k=1 frac1k!$ and from my calculations it should be $leq 2$, but then $eleq 2$ doesn't makes sense :) ... it took me ~10 minutes to realise it's $e-1<2$ in fact.
– rtybase
Sep 9 at 12:07
add a comment |Â
for $displaystyle sum frac1k!=e$ to be true, $k$ needs to start from $0$, otherwise $displaystyle sumlimits_k=1 frac1k!=e-1$
– rtybase
Sep 9 at 11:58
1
Indeed. I nowhere specified where the summation starts, though, I just suggested the idea
– b00n heT
Sep 9 at 12:02
1
The reason I raised it was, for $n=0$ it should lead to $sumlimits_k=1 frac1k!$ and from my calculations it should be $leq 2$, but then $eleq 2$ doesn't makes sense :) ... it took me ~10 minutes to realise it's $e-1<2$ in fact.
– rtybase
Sep 9 at 12:07
for $displaystyle sum frac1k!=e$ to be true, $k$ needs to start from $0$, otherwise $displaystyle sumlimits_k=1 frac1k!=e-1$
– rtybase
Sep 9 at 11:58
for $displaystyle sum frac1k!=e$ to be true, $k$ needs to start from $0$, otherwise $displaystyle sumlimits_k=1 frac1k!=e-1$
– rtybase
Sep 9 at 11:58
1
1
Indeed. I nowhere specified where the summation starts, though, I just suggested the idea
– b00n heT
Sep 9 at 12:02
Indeed. I nowhere specified where the summation starts, though, I just suggested the idea
– b00n heT
Sep 9 at 12:02
1
1
The reason I raised it was, for $n=0$ it should lead to $sumlimits_k=1 frac1k!$ and from my calculations it should be $leq 2$, but then $eleq 2$ doesn't makes sense :) ... it took me ~10 minutes to realise it's $e-1<2$ in fact.
– rtybase
Sep 9 at 12:07
The reason I raised it was, for $n=0$ it should lead to $sumlimits_k=1 frac1k!$ and from my calculations it should be $leq 2$, but then $eleq 2$ doesn't makes sense :) ... it took me ~10 minutes to realise it's $e-1<2$ in fact.
– rtybase
Sep 9 at 12:07
add a comment |Â
up vote
0
down vote
Just focusing on convergence, since $n+kgeq n+2, forall kgeq2,forall ngeq0$ we have:
$$0<frac1(n+1)(n+2)(n+3)...(n+k)leqfrac1(n+1)(n+2)^k-1$$
thus
$$sumlimits_k=1frac1(n+1)(n+2)(n+3)...(n+k)leq
frac1n+1+sumlimits_k=2frac1(n+1)(n+2)^k-1=\
frac1n+1left(1+sumlimits_colorredk=2frac1(n+2)^k-1right)=
frac1n+1left(1+sumlimits_colorredk=1frac1(n+2)^kright)=\
frac1n+1left(sumlimits_colorredk=0frac1(n+2)^kright)=...$$
the latter is an infinite geometric progression, or:
$$...=frac1n+1cdot frac11-frac1n+2=fracn+2(n+1)^2$$
answered Sep 9 at 11:57
rtybase
9,21721433
add a comment |Â
up vote
0
down vote
Just focusing on convergence, since $n+kgeq n+2, forall kgeq2,forall ngeq0$ we have:
$$0<frac1(n+1)(n+2)(n+3)...(n+k)leqfrac1(n+1)(n+2)^k-1$$
thus
$$sumlimits_k=1frac1(n+1)(n+2)(n+3)...(n+k)leq
frac1n+1+sumlimits_k=2frac1(n+1)(n+2)^k-1=\
frac1n+1left(1+sumlimits_colorredk=2frac1(n+2)^k-1right)=
frac1n+1left(1+sumlimits_colorredk=1frac1(n+2)^kright)=\
frac1n+1left(sumlimits_colorredk=0frac1(n+2)^kright)=...$$
the latter is an infinite geometric progression, or:
$$...=frac1n+1cdot frac11-frac1n+2=fracn+2(n+1)^2$$
answered Sep 9 at 11:57
rtybase
9,21721433
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Just focusing on convergence, since $n+kgeq n+2, forall kgeq2,forall ngeq0$ we have:
$$0<frac1(n+1)(n+2)(n+3)...(n+k)leqfrac1(n+1)(n+2)^k-1$$
thus
$$sumlimits_k=1frac1(n+1)(n+2)(n+3)...(n+k)leq
frac1n+1+sumlimits_k=2frac1(n+1)(n+2)^k-1=\
frac1n+1left(1+sumlimits_colorredk=2frac1(n+2)^k-1right)=
frac1n+1left(1+sumlimits_colorredk=1frac1(n+2)^kright)=\
frac1n+1left(sumlimits_colorredk=0frac1(n+2)^kright)=...$$
the latter is an infinite geometric progression, or:
$$...=frac1n+1cdot frac11-frac1n+2=fracn+2(n+1)^2$$
answered Sep 9 at 11:57
rtybase
9,21721433
Just focusing on convergence, since $n+kgeq n+2, forall kgeq2,forall ngeq0$ we have:
$$0<frac1(n+1)(n+2)(n+3)...(n+k)leqfrac1(n+1)(n+2)^k-1$$
thus
$$sumlimits_k=1frac1(n+1)(n+2)(n+3)...(n+k)leq
frac1n+1+sumlimits_k=2frac1(n+1)(n+2)^k-1=\
frac1n+1left(1+sumlimits_colorredk=2frac1(n+2)^k-1right)=
frac1n+1left(1+sumlimits_colorredk=1frac1(n+2)^kright)=\
frac1n+1left(sumlimits_colorredk=0frac1(n+2)^kright)=...$$
the latter is an infinite geometric progression, or:
$$...=frac1n+1cdot frac11-frac1n+2=fracn+2(n+1)^2$$
answered Sep 9 at 11:57
rtybase
9,21721433
edited Sep 9 at 12:03
answered Sep 9 at 11:57
rtybase
9,21721433
answered Sep 9 at 11:57
rtybase
9,21721433
answered Sep 9 at 11:57
rtybase
9,21721433
9,21721433
add a comment |Â
add a comment |Â
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2
Is the sum made over $n$ or over $k$?
– LucaMac
Sep 9 at 10:26
@LucaMac over $k$
– user580093
Sep 9 at 10:27
So it's $sumlimits_k=1^+infty frac1(n+1)cdotldotscdot(n+k)$, right?
– LucaMac
Sep 9 at 10:28