Is this a Riemann sum (if so, I can't figure out which one)?
Clash Royale CLAN TAG#URR8PPP
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This was supposedly an easy limit, and it is suspiciously similar to a Riemann sum, but I can't quite figure out for what function.
$$lim_ntoinftyfrac1n sum_k=3^nfrac3k^2-k-2$$
Well, even the fact that $frac3k^2-k-2 = frac1k-1-frac1k+2$ doesn't seem to simplify the problem. I thought this would be a telescoping sum, but it's clearly not.
Is that a Riemann sum at all?
limits riemann-sum
edited Feb 16 at 19:31
Mutantoe
619513
asked Feb 16 at 14:31
Don DraperDon Draper
87110
$endgroup$
add a comment |
$begingroup$
This was supposedly an easy limit, and it is suspiciously similar to a Riemann sum, but I can't quite figure out for what function.
$$lim_ntoinftyfrac1n sum_k=3^nfrac3k^2-k-2$$
Well, even the fact that $frac3k^2-k-2 = frac1k-1-frac1k+2$ doesn't seem to simplify the problem. I thought this would be a telescoping sum, but it's clearly not.
Is that a Riemann sum at all?
limits riemann-sum
edited Feb 16 at 19:31
Mutantoe
619513
asked Feb 16 at 14:31
Don DraperDon Draper
87110
$endgroup$
$begingroup$
Set $k=3,4,5,6$ etc. and add
$endgroup$
– lab bhattacharjee
Feb 16 at 14:37
$begingroup$
Your partial fraction decomposition does indeed simplify the problem. Lots of terms are cancelled
$endgroup$
– Gabriel Ribeiro
Feb 16 at 14:38
$begingroup$
That is not a Riemann sum. Since the partition width is just $frac 1n$, a similar Riemann sum would have the form $$frac 1n sum fleft(frac knright)$$ for some function $f$. But note that other than the width multiple, the summand depends only on the ratio of $k$ and $n$. This is not the case in your sum.
$endgroup$
– Paul Sinclair
Feb 16 at 20:44
2
$begingroup$
This is an easy limit--it's zero by inspection. Since the sum is clearly convergent , the $1/n$ term dominates the asymptotic behavior.
$endgroup$
– eyeballfrog
Feb 16 at 22:11
add a comment |
$begingroup$
This was supposedly an easy limit, and it is suspiciously similar to a Riemann sum, but I can't quite figure out for what function.
$$lim_ntoinftyfrac1n sum_k=3^nfrac3k^2-k-2$$
Well, even the fact that $frac3k^2-k-2 = frac1k-1-frac1k+2$ doesn't seem to simplify the problem. I thought this would be a telescoping sum, but it's clearly not.
Is that a Riemann sum at all?
limits riemann-sum
edited Feb 16 at 19:31
Mutantoe
619513
asked Feb 16 at 14:31
Don DraperDon Draper
87110
$endgroup$
This was supposedly an easy limit, and it is suspiciously similar to a Riemann sum, but I can't quite figure out for what function.
$$lim_ntoinftyfrac1n sum_k=3^nfrac3k^2-k-2$$
Well, even the fact that $frac3k^2-k-2 = frac1k-1-frac1k+2$ doesn't seem to simplify the problem. I thought this would be a telescoping sum, but it's clearly not.
Is that a Riemann sum at all?
limits riemann-sum
limits riemann-sum
edited Feb 16 at 19:31
Mutantoe
619513
asked Feb 16 at 14:31
Don DraperDon Draper
87110
edited Feb 16 at 19:31
Mutantoe
619513
asked Feb 16 at 14:31
Don DraperDon Draper
87110
edited Feb 16 at 19:31
Mutantoe
619513
edited Feb 16 at 19:31
Mutantoe
619513
edited Feb 16 at 19:31
Mutantoe
619513
619513
asked Feb 16 at 14:31
Don DraperDon Draper
87110
asked Feb 16 at 14:31
Don DraperDon Draper
87110
asked Feb 16 at 14:31
Don DraperDon Draper
87110
87110
$begingroup$
Set $k=3,4,5,6$ etc. and add
$endgroup$
– lab bhattacharjee
Feb 16 at 14:37
$begingroup$
Your partial fraction decomposition does indeed simplify the problem. Lots of terms are cancelled
$endgroup$
– Gabriel Ribeiro
Feb 16 at 14:38
$begingroup$
That is not a Riemann sum. Since the partition width is just $frac 1n$, a similar Riemann sum would have the form $$frac 1n sum fleft(frac knright)$$ for some function $f$. But note that other than the width multiple, the summand depends only on the ratio of $k$ and $n$. This is not the case in your sum.
$endgroup$
– Paul Sinclair
Feb 16 at 20:44
2
$begingroup$
This is an easy limit--it's zero by inspection. Since the sum is clearly convergent , the $1/n$ term dominates the asymptotic behavior.
$endgroup$
– eyeballfrog
Feb 16 at 22:11
add a comment |
$begingroup$
Set $k=3,4,5,6$ etc. and add
$endgroup$
– lab bhattacharjee
Feb 16 at 14:37
$begingroup$
Your partial fraction decomposition does indeed simplify the problem. Lots of terms are cancelled
$endgroup$
– Gabriel Ribeiro
Feb 16 at 14:38
$begingroup$
That is not a Riemann sum. Since the partition width is just $frac 1n$, a similar Riemann sum would have the form $$frac 1n sum fleft(frac knright)$$ for some function $f$. But note that other than the width multiple, the summand depends only on the ratio of $k$ and $n$. This is not the case in your sum.
$endgroup$
– Paul Sinclair
Feb 16 at 20:44
2
$begingroup$
This is an easy limit--it's zero by inspection. Since the sum is clearly convergent , the $1/n$ term dominates the asymptotic behavior.
$endgroup$
– eyeballfrog
Feb 16 at 22:11
$begingroup$
Set $k=3,4,5,6$ etc. and add
$endgroup$
– lab bhattacharjee
Feb 16 at 14:37
$begingroup$
Set $k=3,4,5,6$ etc. and add
$endgroup$
– lab bhattacharjee
Feb 16 at 14:37
$begingroup$
Your partial fraction decomposition does indeed simplify the problem. Lots of terms are cancelled
$endgroup$
– Gabriel Ribeiro
Feb 16 at 14:38
$begingroup$
Your partial fraction decomposition does indeed simplify the problem. Lots of terms are cancelled
$endgroup$
– Gabriel Ribeiro
Feb 16 at 14:38
$begingroup$
That is not a Riemann sum. Since the partition width is just $frac 1n$, a similar Riemann sum would have the form $$frac 1n sum fleft(frac knright)$$ for some function $f$. But note that other than the width multiple, the summand depends only on the ratio of $k$ and $n$. This is not the case in your sum.
$endgroup$
– Paul Sinclair
Feb 16 at 20:44
$begingroup$
That is not a Riemann sum. Since the partition width is just $frac 1n$, a similar Riemann sum would have the form $$frac 1n sum fleft(frac knright)$$ for some function $f$. But note that other than the width multiple, the summand depends only on the ratio of $k$ and $n$. This is not the case in your sum.
$endgroup$
– Paul Sinclair
Feb 16 at 20:44
2
2
$begingroup$
This is an easy limit--it's zero by inspection. Since the sum is clearly convergent , the $1/n$ term dominates the asymptotic behavior.
$endgroup$
– eyeballfrog
Feb 16 at 22:11
$begingroup$
This is an easy limit--it's zero by inspection. Since the sum is clearly convergent , the $1/n$ term dominates the asymptotic behavior.
$endgroup$
– eyeballfrog
Feb 16 at 22:11
add a comment |
2 Answers
2
active
oldest
votes
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$$sum_k=3^n frac3k^2-k-2 = sum_k=3^n frac1k-2- sum_k=3^n frac1k+1$$
You (and I) were mistaken before, see @Romeo 's answer.
Notice that $$sum_k=3^n frac1k-2=sum_k=0^n-3 frac1k+1$$
Insert above you get
$$sum_k=3^n frac3k^2-k-2 = sum_k=0^n-3 frac1k+1 - sum_k=3^n frac1k+1 = sum_k=0^2 frac1k+1 - sum_k=n-2^n frac1k+1= $$$$=1+frac12+frac13 - frac1n-1-frac1n-frac1n+1$$
Of course this argument requires $ngeq 3$.
answered Feb 16 at 14:39
YankoYanko
7,7001830
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add a comment |
$begingroup$
Isn't it $k^2 -k -2 = (k+1)(k-2)$? In that way,
$$
frac3k^2-k-2 = frac1k-2 - frac1k+1
$$
and this is likely to be telescopic.
answered Feb 16 at 14:38
RomeoRomeo
3,06421148
$endgroup$
$begingroup$
It looks like the OP already came up with the equality.
$endgroup$
– Yanko
Feb 16 at 14:39
2
$begingroup$
@Yanko It seems to me the signs of the decomposition of the OP's (and yours) are wrong. Am I mistaken?
$endgroup$
– Romeo
Feb 16 at 14:40
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Nope you are correct!
$endgroup$
– Yanko
Feb 16 at 14:41
1
$begingroup$
@Yanko Thanks a lot, now your post is correct :-)
$endgroup$
– Romeo
Feb 16 at 14:42
add a comment |
Your Answer
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2 Answers
2
active
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2 Answers
2
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oldest
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votes
$begingroup$
$$sum_k=3^n frac3k^2-k-2 = sum_k=3^n frac1k-2- sum_k=3^n frac1k+1$$
You (and I) were mistaken before, see @Romeo 's answer.
Notice that $$sum_k=3^n frac1k-2=sum_k=0^n-3 frac1k+1$$
Insert above you get
$$sum_k=3^n frac3k^2-k-2 = sum_k=0^n-3 frac1k+1 - sum_k=3^n frac1k+1 = sum_k=0^2 frac1k+1 - sum_k=n-2^n frac1k+1= $$$$=1+frac12+frac13 - frac1n-1-frac1n-frac1n+1$$
Of course this argument requires $ngeq 3$.
answered Feb 16 at 14:39
YankoYanko
7,7001830
$endgroup$
add a comment |
$begingroup$
$$sum_k=3^n frac3k^2-k-2 = sum_k=3^n frac1k-2- sum_k=3^n frac1k+1$$
You (and I) were mistaken before, see @Romeo 's answer.
Notice that $$sum_k=3^n frac1k-2=sum_k=0^n-3 frac1k+1$$
Insert above you get
$$sum_k=3^n frac3k^2-k-2 = sum_k=0^n-3 frac1k+1 - sum_k=3^n frac1k+1 = sum_k=0^2 frac1k+1 - sum_k=n-2^n frac1k+1= $$$$=1+frac12+frac13 - frac1n-1-frac1n-frac1n+1$$
Of course this argument requires $ngeq 3$.
answered Feb 16 at 14:39
YankoYanko
7,7001830
$endgroup$
add a comment |
$begingroup$
$$sum_k=3^n frac3k^2-k-2 = sum_k=3^n frac1k-2- sum_k=3^n frac1k+1$$
You (and I) were mistaken before, see @Romeo 's answer.
Notice that $$sum_k=3^n frac1k-2=sum_k=0^n-3 frac1k+1$$
Insert above you get
$$sum_k=3^n frac3k^2-k-2 = sum_k=0^n-3 frac1k+1 - sum_k=3^n frac1k+1 = sum_k=0^2 frac1k+1 - sum_k=n-2^n frac1k+1= $$$$=1+frac12+frac13 - frac1n-1-frac1n-frac1n+1$$
Of course this argument requires $ngeq 3$.
answered Feb 16 at 14:39
YankoYanko
7,7001830
$endgroup$
$$sum_k=3^n frac3k^2-k-2 = sum_k=3^n frac1k-2- sum_k=3^n frac1k+1$$
You (and I) were mistaken before, see @Romeo 's answer.
Notice that $$sum_k=3^n frac1k-2=sum_k=0^n-3 frac1k+1$$
Insert above you get
$$sum_k=3^n frac3k^2-k-2 = sum_k=0^n-3 frac1k+1 - sum_k=3^n frac1k+1 = sum_k=0^2 frac1k+1 - sum_k=n-2^n frac1k+1= $$$$=1+frac12+frac13 - frac1n-1-frac1n-frac1n+1$$
Of course this argument requires $ngeq 3$.
answered Feb 16 at 14:39
YankoYanko
7,7001830
edited Feb 16 at 14:44
answered Feb 16 at 14:39
YankoYanko
7,7001830
answered Feb 16 at 14:39
YankoYanko
7,7001830
answered Feb 16 at 14:39
YankoYanko
7,7001830
7,7001830
add a comment |
add a comment |
$begingroup$
Isn't it $k^2 -k -2 = (k+1)(k-2)$? In that way,
$$
frac3k^2-k-2 = frac1k-2 - frac1k+1
$$
and this is likely to be telescopic.
answered Feb 16 at 14:38
RomeoRomeo
3,06421148
$endgroup$
$begingroup$
It looks like the OP already came up with the equality.
$endgroup$
– Yanko
Feb 16 at 14:39
2
$begingroup$
@Yanko It seems to me the signs of the decomposition of the OP's (and yours) are wrong. Am I mistaken?
$endgroup$
– Romeo
Feb 16 at 14:40
$begingroup$
Nope you are correct!
$endgroup$
– Yanko
Feb 16 at 14:41
1
$begingroup$
@Yanko Thanks a lot, now your post is correct :-)
$endgroup$
– Romeo
Feb 16 at 14:42
add a comment |
$begingroup$
Isn't it $k^2 -k -2 = (k+1)(k-2)$? In that way,
$$
frac3k^2-k-2 = frac1k-2 - frac1k+1
$$
and this is likely to be telescopic.
answered Feb 16 at 14:38
RomeoRomeo
3,06421148
$endgroup$
$begingroup$
It looks like the OP already came up with the equality.
$endgroup$
– Yanko
Feb 16 at 14:39
2
$begingroup$
@Yanko It seems to me the signs of the decomposition of the OP's (and yours) are wrong. Am I mistaken?
$endgroup$
– Romeo
Feb 16 at 14:40
$begingroup$
Nope you are correct!
$endgroup$
– Yanko
Feb 16 at 14:41
1
$begingroup$
@Yanko Thanks a lot, now your post is correct :-)
$endgroup$
– Romeo
Feb 16 at 14:42
add a comment |
$begingroup$
Isn't it $k^2 -k -2 = (k+1)(k-2)$? In that way,
$$
frac3k^2-k-2 = frac1k-2 - frac1k+1
$$
and this is likely to be telescopic.
answered Feb 16 at 14:38
RomeoRomeo
3,06421148
$endgroup$
Isn't it $k^2 -k -2 = (k+1)(k-2)$? In that way,
$$
frac3k^2-k-2 = frac1k-2 - frac1k+1
$$
and this is likely to be telescopic.
answered Feb 16 at 14:38
RomeoRomeo
3,06421148
answered Feb 16 at 14:38
RomeoRomeo
3,06421148
answered Feb 16 at 14:38
RomeoRomeo
3,06421148
answered Feb 16 at 14:38
RomeoRomeo
3,06421148
3,06421148
$begingroup$
It looks like the OP already came up with the equality.
$endgroup$
– Yanko
Feb 16 at 14:39
2
$begingroup$
@Yanko It seems to me the signs of the decomposition of the OP's (and yours) are wrong. Am I mistaken?
$endgroup$
– Romeo
Feb 16 at 14:40
$begingroup$
Nope you are correct!
$endgroup$
– Yanko
Feb 16 at 14:41
1
$begingroup$
@Yanko Thanks a lot, now your post is correct :-)
$endgroup$
– Romeo
Feb 16 at 14:42
add a comment |
$begingroup$
It looks like the OP already came up with the equality.
$endgroup$
– Yanko
Feb 16 at 14:39
2
$begingroup$
@Yanko It seems to me the signs of the decomposition of the OP's (and yours) are wrong. Am I mistaken?
$endgroup$
– Romeo
Feb 16 at 14:40
$begingroup$
Nope you are correct!
$endgroup$
– Yanko
Feb 16 at 14:41
1
$begingroup$
@Yanko Thanks a lot, now your post is correct :-)
$endgroup$
– Romeo
Feb 16 at 14:42
$begingroup$
It looks like the OP already came up with the equality.
$endgroup$
– Yanko
Feb 16 at 14:39
$begingroup$
It looks like the OP already came up with the equality.
$endgroup$
– Yanko
Feb 16 at 14:39
2
2
$begingroup$
@Yanko It seems to me the signs of the decomposition of the OP's (and yours) are wrong. Am I mistaken?
$endgroup$
– Romeo
Feb 16 at 14:40
$begingroup$
@Yanko It seems to me the signs of the decomposition of the OP's (and yours) are wrong. Am I mistaken?
$endgroup$
– Romeo
Feb 16 at 14:40
$begingroup$
Nope you are correct!
$endgroup$
– Yanko
Feb 16 at 14:41
$begingroup$
Nope you are correct!
$endgroup$
– Yanko
Feb 16 at 14:41
1
1
$begingroup$
@Yanko Thanks a lot, now your post is correct :-)
$endgroup$
– Romeo
Feb 16 at 14:42
$begingroup$
@Yanko Thanks a lot, now your post is correct :-)
$endgroup$
– Romeo
Feb 16 at 14:42
add a comment |
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$begingroup$
Set $k=3,4,5,6$ etc. and add
$endgroup$
– lab bhattacharjee
Feb 16 at 14:37
$begingroup$
Your partial fraction decomposition does indeed simplify the problem. Lots of terms are cancelled
$endgroup$
– Gabriel Ribeiro
Feb 16 at 14:38
$begingroup$
That is not a Riemann sum. Since the partition width is just $frac 1n$, a similar Riemann sum would have the form $$frac 1n sum fleft(frac knright)$$ for some function $f$. But note that other than the width multiple, the summand depends only on the ratio of $k$ and $n$. This is not the case in your sum.
$endgroup$
– Paul Sinclair
Feb 16 at 20:44
2
$begingroup$
This is an easy limit--it's zero by inspection. Since the sum is clearly convergent , the $1/n$ term dominates the asymptotic behavior.
$endgroup$
– eyeballfrog
Feb 16 at 22:11