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Find an equivalent to $sum_k=n^infty frac1k!$
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I would like to find an equivalent of $sum_k=n^infty frac1k!$. Can I do as follow ? (since I always have doubt with those $o$ and $O$, I would like your opinion.
$$n!sum_k=n^infty frac1k!=1+frac1n+1+frac1(n+1)(n+2)+frac1(n+1)(n+2)(n+3)+cdots$$
$$=1+frac1n+1+mathcal Oleft(sum_k=n+1^infty frac1k^2right)=1+o(1).$$
and thus the claim follow.
Q1) Is it correct ?
Q2)
I have that $mathcal Oleft(sum_k=n+1^infty frac1k^2right)$ because $$frac1(n+1)(n+2)leq frac1(n+1)^2,quad frac1(n+1)(n+2)(n+3)leq frac1(n+2)^2,quad frac1(n+1)(n+2)(n+3)(n+4)leqfrac1(n+3)^2 cdots$$
and so $frac1(n+1)(n+2)cdots(n+k)leq frac1(n+k)^2$. But suppose we could only prove that $$frac1(n+1)(n+2)cdots(n+k)leq fracC_k(n+k)^2,$$
i.e. that the constant depend on $k$ and we can't have a uniform bound. Could I conclude that $$ frac1(n+1)(n+2)+frac1(n+1)(n+2)(n+3)+cdots=mathcal Oleft(sum_k=n+1^infty frac1k^2right) ?$$
asymptotics
edited Jul 30 at 20:26
Michael Hardy
204k23185460
asked Jul 30 at 20:20
Peter
326112
add a comment |Â
up vote
1
down vote
favorite
I would like to find an equivalent of $sum_k=n^infty frac1k!$. Can I do as follow ? (since I always have doubt with those $o$ and $O$, I would like your opinion.
$$n!sum_k=n^infty frac1k!=1+frac1n+1+frac1(n+1)(n+2)+frac1(n+1)(n+2)(n+3)+cdots$$
$$=1+frac1n+1+mathcal Oleft(sum_k=n+1^infty frac1k^2right)=1+o(1).$$
and thus the claim follow.
Q1) Is it correct ?
Q2)
I have that $mathcal Oleft(sum_k=n+1^infty frac1k^2right)$ because $$frac1(n+1)(n+2)leq frac1(n+1)^2,quad frac1(n+1)(n+2)(n+3)leq frac1(n+2)^2,quad frac1(n+1)(n+2)(n+3)(n+4)leqfrac1(n+3)^2 cdots$$
and so $frac1(n+1)(n+2)cdots(n+k)leq frac1(n+k)^2$. But suppose we could only prove that $$frac1(n+1)(n+2)cdots(n+k)leq fracC_k(n+k)^2,$$
i.e. that the constant depend on $k$ and we can't have a uniform bound. Could I conclude that $$ frac1(n+1)(n+2)+frac1(n+1)(n+2)(n+3)+cdots=mathcal Oleft(sum_k=n+1^infty frac1k^2right) ?$$
asymptotics
edited Jul 30 at 20:26
Michael Hardy
204k23185460
asked Jul 30 at 20:20
Peter
326112
What do you mean by "find an equivalent to"? It seems to me that the expression is $frac(n-1)!e(n-1)!$.
– A. Pongrácz
Jul 30 at 20:28
an equivalent of $u_n$ is a sequence $v_n$ s.t. $u_n/v_nto 1$. @A.Pongrácz
– Peter
Jul 30 at 20:30
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I would like to find an equivalent of $sum_k=n^infty frac1k!$. Can I do as follow ? (since I always have doubt with those $o$ and $O$, I would like your opinion.
$$n!sum_k=n^infty frac1k!=1+frac1n+1+frac1(n+1)(n+2)+frac1(n+1)(n+2)(n+3)+cdots$$
$$=1+frac1n+1+mathcal Oleft(sum_k=n+1^infty frac1k^2right)=1+o(1).$$
and thus the claim follow.
Q1) Is it correct ?
Q2)
I have that $mathcal Oleft(sum_k=n+1^infty frac1k^2right)$ because $$frac1(n+1)(n+2)leq frac1(n+1)^2,quad frac1(n+1)(n+2)(n+3)leq frac1(n+2)^2,quad frac1(n+1)(n+2)(n+3)(n+4)leqfrac1(n+3)^2 cdots$$
and so $frac1(n+1)(n+2)cdots(n+k)leq frac1(n+k)^2$. But suppose we could only prove that $$frac1(n+1)(n+2)cdots(n+k)leq fracC_k(n+k)^2,$$
i.e. that the constant depend on $k$ and we can't have a uniform bound. Could I conclude that $$ frac1(n+1)(n+2)+frac1(n+1)(n+2)(n+3)+cdots=mathcal Oleft(sum_k=n+1^infty frac1k^2right) ?$$
asymptotics
edited Jul 30 at 20:26
Michael Hardy
204k23185460
asked Jul 30 at 20:20
Peter
326112
I would like to find an equivalent of $sum_k=n^infty frac1k!$. Can I do as follow ? (since I always have doubt with those $o$ and $O$, I would like your opinion.
$$n!sum_k=n^infty frac1k!=1+frac1n+1+frac1(n+1)(n+2)+frac1(n+1)(n+2)(n+3)+cdots$$
$$=1+frac1n+1+mathcal Oleft(sum_k=n+1^infty frac1k^2right)=1+o(1).$$
and thus the claim follow.
Q1) Is it correct ?
Q2)
I have that $mathcal Oleft(sum_k=n+1^infty frac1k^2right)$ because $$frac1(n+1)(n+2)leq frac1(n+1)^2,quad frac1(n+1)(n+2)(n+3)leq frac1(n+2)^2,quad frac1(n+1)(n+2)(n+3)(n+4)leqfrac1(n+3)^2 cdots$$
and so $frac1(n+1)(n+2)cdots(n+k)leq frac1(n+k)^2$. But suppose we could only prove that $$frac1(n+1)(n+2)cdots(n+k)leq fracC_k(n+k)^2,$$
i.e. that the constant depend on $k$ and we can't have a uniform bound. Could I conclude that $$ frac1(n+1)(n+2)+frac1(n+1)(n+2)(n+3)+cdots=mathcal Oleft(sum_k=n+1^infty frac1k^2right) ?$$
asymptotics
edited Jul 30 at 20:26
Michael Hardy
204k23185460
asked Jul 30 at 20:20
Peter
326112
edited Jul 30 at 20:26
Michael Hardy
204k23185460
edited Jul 30 at 20:26
Michael Hardy
204k23185460
edited Jul 30 at 20:26
Michael Hardy
204k23185460
204k23185460
asked Jul 30 at 20:20
Peter
326112
asked Jul 30 at 20:20
Peter
326112
asked Jul 30 at 20:20
Peter
326112
326112
What do you mean by "find an equivalent to"? It seems to me that the expression is $frac(n-1)!e(n-1)!$.
– A. Pongrácz
Jul 30 at 20:28
an equivalent of $u_n$ is a sequence $v_n$ s.t. $u_n/v_nto 1$. @A.Pongrácz
– Peter
Jul 30 at 20:30
add a comment |Â
What do you mean by "find an equivalent to"? It seems to me that the expression is $frac(n-1)!e(n-1)!$.
– A. Pongrácz
Jul 30 at 20:28
an equivalent of $u_n$ is a sequence $v_n$ s.t. $u_n/v_nto 1$. @A.Pongrácz
– Peter
Jul 30 at 20:30
What do you mean by "find an equivalent to"? It seems to me that the expression is $frac(n-1)!e(n-1)!$.
– A. Pongrácz
Jul 30 at 20:28
What do you mean by "find an equivalent to"? It seems to me that the expression is $frac(n-1)!e(n-1)!$.
– A. Pongrácz
Jul 30 at 20:28
an equivalent of $u_n$ is a sequence $v_n$ s.t. $u_n/v_nto 1$. @A.Pongrácz
– Peter
Jul 30 at 20:30
an equivalent of $u_n$ is a sequence $v_n$ s.t. $u_n/v_nto 1$. @A.Pongrácz
– Peter
Jul 30 at 20:30
add a comment |Â
1 Answer
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-1
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See this post: Calculate $lim_n rightarrow infty$ ($n!e-[n!e]$)?
It follows that the integer part of $(n-1)!e$ is $(n-1)!sumlimits_k=1^n-1 frac1k!$, so the fractional part of $(n-1)!e$ is $(n-1)!sumlimits_k=n^infty frac1k!$.
Hence, an equivalent expression is $frac(n-1)!e(n-1)!$.
answered Jul 31 at 5:17
A. Pongrácz
1,309115
Sorry but it doesn't answer my question. I know it's equivalent. I just want to know if my method is correct, and if my Q2) is right.
– Peter
Jul 31 at 7:25
Actually, it does. If you check the post I linked, the same calcualtion is also carried out there, showing that your calculation is correct. Your question was not specific. You answer was that $frac1n!$ is asymptotically equal to the expression, mine was that $frac(n-1)!e(n-1)!$ is EQUAL to the expression. As the question was not concrete, anyone can pick their favourite answer, and I prefer mine to yours. With that being said, yours was also correct.
– A. Pongrácz
Jul 31 at 9:56
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
-1
down vote
See this post: Calculate $lim_n rightarrow infty$ ($n!e-[n!e]$)?
It follows that the integer part of $(n-1)!e$ is $(n-1)!sumlimits_k=1^n-1 frac1k!$, so the fractional part of $(n-1)!e$ is $(n-1)!sumlimits_k=n^infty frac1k!$.
Hence, an equivalent expression is $frac(n-1)!e(n-1)!$.
answered Jul 31 at 5:17
A. Pongrácz
1,309115
Sorry but it doesn't answer my question. I know it's equivalent. I just want to know if my method is correct, and if my Q2) is right.
– Peter
Jul 31 at 7:25
Actually, it does. If you check the post I linked, the same calcualtion is also carried out there, showing that your calculation is correct. Your question was not specific. You answer was that $frac1n!$ is asymptotically equal to the expression, mine was that $frac(n-1)!e(n-1)!$ is EQUAL to the expression. As the question was not concrete, anyone can pick their favourite answer, and I prefer mine to yours. With that being said, yours was also correct.
– A. Pongrácz
Jul 31 at 9:56
add a comment |Â
up vote
-1
down vote
See this post: Calculate $lim_n rightarrow infty$ ($n!e-[n!e]$)?
It follows that the integer part of $(n-1)!e$ is $(n-1)!sumlimits_k=1^n-1 frac1k!$, so the fractional part of $(n-1)!e$ is $(n-1)!sumlimits_k=n^infty frac1k!$.
Hence, an equivalent expression is $frac(n-1)!e(n-1)!$.
answered Jul 31 at 5:17
A. Pongrácz
1,309115
Sorry but it doesn't answer my question. I know it's equivalent. I just want to know if my method is correct, and if my Q2) is right.
– Peter
Jul 31 at 7:25
Actually, it does. If you check the post I linked, the same calcualtion is also carried out there, showing that your calculation is correct. Your question was not specific. You answer was that $frac1n!$ is asymptotically equal to the expression, mine was that $frac(n-1)!e(n-1)!$ is EQUAL to the expression. As the question was not concrete, anyone can pick their favourite answer, and I prefer mine to yours. With that being said, yours was also correct.
– A. Pongrácz
Jul 31 at 9:56
add a comment |Â
up vote
-1
down vote
up vote
-1
down vote
See this post: Calculate $lim_n rightarrow infty$ ($n!e-[n!e]$)?
It follows that the integer part of $(n-1)!e$ is $(n-1)!sumlimits_k=1^n-1 frac1k!$, so the fractional part of $(n-1)!e$ is $(n-1)!sumlimits_k=n^infty frac1k!$.
Hence, an equivalent expression is $frac(n-1)!e(n-1)!$.
answered Jul 31 at 5:17
A. Pongrácz
1,309115
See this post: Calculate $lim_n rightarrow infty$ ($n!e-[n!e]$)?
It follows that the integer part of $(n-1)!e$ is $(n-1)!sumlimits_k=1^n-1 frac1k!$, so the fractional part of $(n-1)!e$ is $(n-1)!sumlimits_k=n^infty frac1k!$.
Hence, an equivalent expression is $frac(n-1)!e(n-1)!$.
answered Jul 31 at 5:17
A. Pongrácz
1,309115
answered Jul 31 at 5:17
A. Pongrácz
1,309115
answered Jul 31 at 5:17
A. Pongrácz
1,309115
answered Jul 31 at 5:17
A. Pongrácz
1,309115
1,309115
Sorry but it doesn't answer my question. I know it's equivalent. I just want to know if my method is correct, and if my Q2) is right.
– Peter
Jul 31 at 7:25
Actually, it does. If you check the post I linked, the same calcualtion is also carried out there, showing that your calculation is correct. Your question was not specific. You answer was that $frac1n!$ is asymptotically equal to the expression, mine was that $frac(n-1)!e(n-1)!$ is EQUAL to the expression. As the question was not concrete, anyone can pick their favourite answer, and I prefer mine to yours. With that being said, yours was also correct.
– A. Pongrácz
Jul 31 at 9:56
add a comment |Â
Sorry but it doesn't answer my question. I know it's equivalent. I just want to know if my method is correct, and if my Q2) is right.
– Peter
Jul 31 at 7:25
Actually, it does. If you check the post I linked, the same calcualtion is also carried out there, showing that your calculation is correct. Your question was not specific. You answer was that $frac1n!$ is asymptotically equal to the expression, mine was that $frac(n-1)!e(n-1)!$ is EQUAL to the expression. As the question was not concrete, anyone can pick their favourite answer, and I prefer mine to yours. With that being said, yours was also correct.
– A. Pongrácz
Jul 31 at 9:56
Sorry but it doesn't answer my question. I know it's equivalent. I just want to know if my method is correct, and if my Q2) is right.
– Peter
Jul 31 at 7:25
Sorry but it doesn't answer my question. I know it's equivalent. I just want to know if my method is correct, and if my Q2) is right.
– Peter
Jul 31 at 7:25
Actually, it does. If you check the post I linked, the same calcualtion is also carried out there, showing that your calculation is correct. Your question was not specific. You answer was that $frac1n!$ is asymptotically equal to the expression, mine was that $frac(n-1)!e(n-1)!$ is EQUAL to the expression. As the question was not concrete, anyone can pick their favourite answer, and I prefer mine to yours. With that being said, yours was also correct.
– A. Pongrácz
Jul 31 at 9:56
Actually, it does. If you check the post I linked, the same calcualtion is also carried out there, showing that your calculation is correct. Your question was not specific. You answer was that $frac1n!$ is asymptotically equal to the expression, mine was that $frac(n-1)!e(n-1)!$ is EQUAL to the expression. As the question was not concrete, anyone can pick their favourite answer, and I prefer mine to yours. With that being said, yours was also correct.
– A. Pongrácz
Jul 31 at 9:56
add a comment |Â
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What do you mean by "find an equivalent to"? It seems to me that the expression is $frac(n-1)!e(n-1)!$.
– A. Pongrácz
Jul 30 at 20:28
an equivalent of $u_n$ is a sequence $v_n$ s.t. $u_n/v_nto 1$. @A.Pongrácz
– Peter
Jul 30 at 20:30