$sum_{n = 1}^{D - 1} frac{n}{D - n}$ written as a function of $D$? [closed]
$$sum_{n = 1}^{D - 1} frac{n}{D - n}$$
Is it possible to reduce this summation to a function of $D$?
summation
edited Nov 28 at 12:20
amWhy
191k28224439
asked Nov 28 at 12:05
bestscammer5
1
closed as unclear what you're asking by amWhy, Saad, Alexander Gruber♦ Nov 30 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.
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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 at 12:20
amWhy
191k28224439
asked Nov 28 at 12:05
bestscammer5
1
closed as unclear what you're asking by amWhy, Saad, Alexander Gruber♦ Nov 30 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
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?
– Alex Vong
Nov 28 at 12:22
add a comment |
$$sum_{n = 1}^{D - 1} frac{n}{D - n}$$
Is it possible to reduce this summation to a function of $D$?
summation
edited Nov 28 at 12:20
amWhy
191k28224439
asked Nov 28 at 12:05
bestscammer5
1
$$sum_{n = 1}^{D - 1} frac{n}{D - n}$$
Is it possible to reduce this summation to a function of $D$?
summation
summation
edited Nov 28 at 12:20
amWhy
191k28224439
asked Nov 28 at 12:05
bestscammer5
1
edited Nov 28 at 12:20
amWhy
191k28224439
asked Nov 28 at 12:05
bestscammer5
1
edited Nov 28 at 12:20
amWhy
191k28224439
edited Nov 28 at 12:20
amWhy
191k28224439
edited Nov 28 at 12:20
amWhy
191k28224439
191k28224439
asked Nov 28 at 12:05
bestscammer5
1
asked Nov 28 at 12:05
bestscammer5
1
asked Nov 28 at 12:05
bestscammer5
1
1
closed as unclear what you're asking by amWhy, Saad, Alexander Gruber♦ Nov 30 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 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
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?
– Alex Vong
Nov 28 at 12:22
add a comment |
3
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?
– Alex Vong
Nov 28 at 12:22
3
3
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?
– Alex Vong
Nov 28 at 12:22
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?
– Alex Vong
Nov 28 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 at 12:33
Calvin Khor
11.2k21438
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$$sum_{n = 1}^{D - 1} frac{n}{D - n}\
=sum_{n=1}^{D-1}left( -1 + frac{D}{D-n}right)\
=-D+1 + Dsum_{n=1}^{D-1}frac1{D-n} \
=-D+1 + Dsum_{m=1}^{D-1}frac1{m} \
= -D+1 + D H_{D-1}$$
where $H_n$ is the $n$th harmonic number.
answered Nov 28 at 12:33
Calvin Khor
11.2k21438
add a comment |
$$sum_{n = 1}^{D - 1} frac{n}{D - n}\
=sum_{n=1}^{D-1}left( -1 + frac{D}{D-n}right)\
=-D+1 + Dsum_{n=1}^{D-1}frac1{D-n} \
=-D+1 + Dsum_{m=1}^{D-1}frac1{m} \
= -D+1 + D H_{D-1}$$
where $H_n$ is the $n$th harmonic number.
answered Nov 28 at 12:33
Calvin Khor
11.2k21438
add a comment |
$$sum_{n = 1}^{D - 1} frac{n}{D - n}\
=sum_{n=1}^{D-1}left( -1 + frac{D}{D-n}right)\
=-D+1 + Dsum_{n=1}^{D-1}frac1{D-n} \
=-D+1 + Dsum_{m=1}^{D-1}frac1{m} \
= -D+1 + D H_{D-1}$$
where $H_n$ is the $n$th harmonic number.
answered Nov 28 at 12:33
Calvin Khor
11.2k21438
$$sum_{n = 1}^{D - 1} frac{n}{D - n}\
=sum_{n=1}^{D-1}left( -1 + frac{D}{D-n}right)\
=-D+1 + Dsum_{n=1}^{D-1}frac1{D-n} \
=-D+1 + Dsum_{m=1}^{D-1}frac1{m} \
= -D+1 + D H_{D-1}$$
where $H_n$ is the $n$th harmonic number.
answered Nov 28 at 12:33
Calvin Khor
11.2k21438
answered Nov 28 at 12:33
Calvin Khor
11.2k21438
answered Nov 28 at 12:33
Calvin Khor
11.2k21438
answered Nov 28 at 12:33
Calvin Khor
11.2k21438
11.2k21438
add a comment |
add a comment |
3
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?
– Alex Vong
Nov 28 at 12:22