$sum_{n = 1}^{D - 1} frac{n}{D - n}$ written as a function of $D$? [closed]
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$$sum_{n = 1}^{D - 1} frac{n}{D - n}$$
Is it possible to reduce this summation to a function of $D$?
summation
edited Nov 28 '18 at 12:20
amWhy
1
asked Nov 28 '18 at 12:05
bestscammer5bestscammer5
1
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closed as unclear what you're asking by amWhy, Saad, Alexander Gruber♦ Nov 30 '18 at 3:36
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
$$sum_{n = 1}^{D - 1} frac{n}{D - n}$$
Is it possible to reduce this summation to a function of $D$?
summation
edited Nov 28 '18 at 12:20
amWhy
1
asked Nov 28 '18 at 12:05
bestscammer5bestscammer5
1
$endgroup$
closed as unclear what you're asking by amWhy, Saad, Alexander Gruber♦ Nov 30 '18 at 3:36
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
3
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Let $$f(D) = sum_{n = 1}^{D - 1} frac{n}{D - n}$$ Then $f$ is already a function of $D$. Are you asking for a closed form for the summation?
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– Alex Vong
Nov 28 '18 at 12:22
add a comment |
$begingroup$
$$sum_{n = 1}^{D - 1} frac{n}{D - n}$$
Is it possible to reduce this summation to a function of $D$?
summation
edited Nov 28 '18 at 12:20
amWhy
1
asked Nov 28 '18 at 12:05
bestscammer5bestscammer5
1
$endgroup$
$$sum_{n = 1}^{D - 1} frac{n}{D - n}$$
Is it possible to reduce this summation to a function of $D$?
summation
summation
edited Nov 28 '18 at 12:20
amWhy
1
asked Nov 28 '18 at 12:05
bestscammer5bestscammer5
1
edited Nov 28 '18 at 12:20
amWhy
1
asked Nov 28 '18 at 12:05
bestscammer5bestscammer5
1
edited Nov 28 '18 at 12:20
amWhy
1
edited Nov 28 '18 at 12:20
amWhy
1
edited Nov 28 '18 at 12:20
amWhy
1
1
asked Nov 28 '18 at 12:05
bestscammer5bestscammer5
1
asked Nov 28 '18 at 12:05
bestscammer5bestscammer5
1
asked Nov 28 '18 at 12:05
bestscammer5bestscammer5
1
1
closed as unclear what you're asking by amWhy, Saad, Alexander Gruber♦ Nov 30 '18 at 3:36
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by amWhy, Saad, Alexander Gruber♦ Nov 30 '18 at 3:36
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
Let $$f(D) = sum_{n = 1}^{D - 1} frac{n}{D - n}$$ Then $f$ is already a function of $D$. Are you asking for a closed form for the summation?
$endgroup$
– Alex Vong
Nov 28 '18 at 12:22
add a comment |
3
$begingroup$
Let $$f(D) = sum_{n = 1}^{D - 1} frac{n}{D - n}$$ Then $f$ is already a function of $D$. Are you asking for a closed form for the summation?
$endgroup$
– Alex Vong
Nov 28 '18 at 12:22
3
3
$begingroup$
Let $$f(D) = sum_{n = 1}^{D - 1} frac{n}{D - n}$$ Then $f$ is already a function of $D$. Are you asking for a closed form for the summation?
$endgroup$
– Alex Vong
Nov 28 '18 at 12:22
$begingroup$
Let $$f(D) = sum_{n = 1}^{D - 1} frac{n}{D - n}$$ Then $f$ is already a function of $D$. Are you asking for a closed form for the summation?
$endgroup$
– Alex Vong
Nov 28 '18 at 12:22
add a comment |
1 Answer
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$$sum_{n = 1}^{D - 1} frac{n}{D - n}\
=sum_{n=1}^{D-1}left( -1 + frac{D}{D-n}right)\
=-D+1 + Dsum_{n=1}^{D-1}frac1{D-n} \
=-D+1 + Dsum_{m=1}^{D-1}frac1{m} \
= -D+1 + D H_{D-1}$$
where $H_n$ is the $n$th harmonic number.
answered Nov 28 '18 at 12:33
Calvin KhorCalvin Khor
11.6k21438
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$$sum_{n = 1}^{D - 1} frac{n}{D - n}\
=sum_{n=1}^{D-1}left( -1 + frac{D}{D-n}right)\
=-D+1 + Dsum_{n=1}^{D-1}frac1{D-n} \
=-D+1 + Dsum_{m=1}^{D-1}frac1{m} \
= -D+1 + D H_{D-1}$$
where $H_n$ is the $n$th harmonic number.
answered Nov 28 '18 at 12:33
Calvin KhorCalvin Khor
11.6k21438
$endgroup$
add a comment |
$begingroup$
$$sum_{n = 1}^{D - 1} frac{n}{D - n}\
=sum_{n=1}^{D-1}left( -1 + frac{D}{D-n}right)\
=-D+1 + Dsum_{n=1}^{D-1}frac1{D-n} \
=-D+1 + Dsum_{m=1}^{D-1}frac1{m} \
= -D+1 + D H_{D-1}$$
where $H_n$ is the $n$th harmonic number.
answered Nov 28 '18 at 12:33
Calvin KhorCalvin Khor
11.6k21438
$endgroup$
add a comment |
$begingroup$
$$sum_{n = 1}^{D - 1} frac{n}{D - n}\
=sum_{n=1}^{D-1}left( -1 + frac{D}{D-n}right)\
=-D+1 + Dsum_{n=1}^{D-1}frac1{D-n} \
=-D+1 + Dsum_{m=1}^{D-1}frac1{m} \
= -D+1 + D H_{D-1}$$
where $H_n$ is the $n$th harmonic number.
answered Nov 28 '18 at 12:33
Calvin KhorCalvin Khor
11.6k21438
$endgroup$
$$sum_{n = 1}^{D - 1} frac{n}{D - n}\
=sum_{n=1}^{D-1}left( -1 + frac{D}{D-n}right)\
=-D+1 + Dsum_{n=1}^{D-1}frac1{D-n} \
=-D+1 + Dsum_{m=1}^{D-1}frac1{m} \
= -D+1 + D H_{D-1}$$
where $H_n$ is the $n$th harmonic number.
answered Nov 28 '18 at 12:33
Calvin KhorCalvin Khor
11.6k21438
answered Nov 28 '18 at 12:33
Calvin KhorCalvin Khor
11.6k21438
answered Nov 28 '18 at 12:33
Calvin KhorCalvin Khor
11.6k21438
answered Nov 28 '18 at 12:33
Calvin KhorCalvin Khor
11.6k21438
11.6k21438
add a comment |
add a comment |
3
$begingroup$
Let $$f(D) = sum_{n = 1}^{D - 1} frac{n}{D - n}$$ Then $f$ is already a function of $D$. Are you asking for a closed form for the summation?
$endgroup$
– Alex Vong
Nov 28 '18 at 12:22