sum of reciprocal numbers of combinations
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Let $^nC_k:=dfrac{n!}{k!(n-k)!}$
Please prove that,for all natural number $k≥2$, $displaystylesum_{n＝k+1}^{infty}frac{1}{^nC_k}＝frac{1}{k-1}$
I tried to prove by induction, but I cannot. I guess it is proved by using Tayler series for some function, but I cannot find the function.
combinatorics taylor-expansion
edited Nov 17 at 14:56
Yadati Kiran
1,239317
asked Nov 17 at 14:44
J.Rie
183
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up vote
2
down vote
favorite
Let $^nC_k:=dfrac{n!}{k!(n-k)!}$
Please prove that,for all natural number $k≥2$, $displaystylesum_{n＝k+1}^{infty}frac{1}{^nC_k}＝frac{1}{k-1}$
I tried to prove by induction, but I cannot. I guess it is proved by using Tayler series for some function, but I cannot find the function.
combinatorics taylor-expansion
edited Nov 17 at 14:56
Yadati Kiran
1,239317
asked Nov 17 at 14:44
J.Rie
183
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $^nC_k:=dfrac{n!}{k!(n-k)!}$
Please prove that,for all natural number $k≥2$, $displaystylesum_{n＝k+1}^{infty}frac{1}{^nC_k}＝frac{1}{k-1}$
I tried to prove by induction, but I cannot. I guess it is proved by using Tayler series for some function, but I cannot find the function.
combinatorics taylor-expansion
edited Nov 17 at 14:56
Yadati Kiran
1,239317
asked Nov 17 at 14:44
J.Rie
183
Let $^nC_k:=dfrac{n!}{k!(n-k)!}$
Please prove that,for all natural number $k≥2$, $displaystylesum_{n＝k+1}^{infty}frac{1}{^nC_k}＝frac{1}{k-1}$
I tried to prove by induction, but I cannot. I guess it is proved by using Tayler series for some function, but I cannot find the function.
combinatorics taylor-expansion
combinatorics taylor-expansion
edited Nov 17 at 14:56
Yadati Kiran
1,239317
asked Nov 17 at 14:44
J.Rie
183
edited Nov 17 at 14:56
Yadati Kiran
1,239317
asked Nov 17 at 14:44
J.Rie
183
edited Nov 17 at 14:56
Yadati Kiran
1,239317
edited Nov 17 at 14:56
Yadati Kiran
1,239317
edited Nov 17 at 14:56
Yadati Kiran
1,239317
1,239317
asked Nov 17 at 14:44
J.Rie
183
asked Nov 17 at 14:44
J.Rie
183
asked Nov 17 at 14:44
J.Rie
183
183
add a comment |
add a comment |
2 Answers
2
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oldest
votes
up vote
1
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accepted
That is known as the German tank problem, and is one of the fundamental Binomial Identities.
answered Nov 17 at 15:54
G Cab
17.1k31237
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0
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This is very simple to prove. All you need to note is the following identity :
$$
frac 1{a(a+1)...(a+m)} = frac{1}{m}left( frac 1{a(a+1)...(a+m-1)} - frac 1{(a+1)...(a+m)}right)
$$
Now, consider the sum:
$$
sum_{n=k+1}^{k+m} frac 1{binom nk} = k! times sum_{i=1}^{i=m} frac{i!}{(k+i)!} \ = k! sum_{i=1}^{i=m}frac 1{(i+1)(i+2)...(i+k)} \ = k! sum_{i=1}^{i=m} frac{1}{k-1}left(frac{1}{(i+1)...(i+k-1)} - frac 1{(i+2)...(i+k)}right) \ = frac{k!}{k-1}sum_{i=1}^{i=m} left(frac{1}{(i+1)...(i+k-1)} - frac 1{(i+2)...(i+k)}right) \ = frac{k!}{k-1} left(frac 1{k!} - frac 1{(m+2)...(m+k)}right) \ = frac 1{k-1} - frac{k!}{(k-1)(m+2)...(m+k)}
$$
By letting $m$ go to infinity, the answer is clearly $frac 1{k-1}$, since the second term goes to zero.
answered Nov 17 at 16:13
астон вілла олоф мэллбэрг
36.7k33376
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
That is known as the German tank problem, and is one of the fundamental Binomial Identities.
answered Nov 17 at 15:54
G Cab
17.1k31237
add a comment |
up vote
1
down vote
accepted
That is known as the German tank problem, and is one of the fundamental Binomial Identities.
answered Nov 17 at 15:54
G Cab
17.1k31237
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
That is known as the German tank problem, and is one of the fundamental Binomial Identities.
answered Nov 17 at 15:54
G Cab
17.1k31237
That is known as the German tank problem, and is one of the fundamental Binomial Identities.
answered Nov 17 at 15:54
G Cab
17.1k31237
answered Nov 17 at 15:54
G Cab
17.1k31237
answered Nov 17 at 15:54
G Cab
17.1k31237
answered Nov 17 at 15:54
G Cab
17.1k31237
17.1k31237
add a comment |
add a comment |
up vote
0
down vote
This is very simple to prove. All you need to note is the following identity :
$$
frac 1{a(a+1)...(a+m)} = frac{1}{m}left( frac 1{a(a+1)...(a+m-1)} - frac 1{(a+1)...(a+m)}right)
$$
Now, consider the sum:
$$
sum_{n=k+1}^{k+m} frac 1{binom nk} = k! times sum_{i=1}^{i=m} frac{i!}{(k+i)!} \ = k! sum_{i=1}^{i=m}frac 1{(i+1)(i+2)...(i+k)} \ = k! sum_{i=1}^{i=m} frac{1}{k-1}left(frac{1}{(i+1)...(i+k-1)} - frac 1{(i+2)...(i+k)}right) \ = frac{k!}{k-1}sum_{i=1}^{i=m} left(frac{1}{(i+1)...(i+k-1)} - frac 1{(i+2)...(i+k)}right) \ = frac{k!}{k-1} left(frac 1{k!} - frac 1{(m+2)...(m+k)}right) \ = frac 1{k-1} - frac{k!}{(k-1)(m+2)...(m+k)}
$$
By letting $m$ go to infinity, the answer is clearly $frac 1{k-1}$, since the second term goes to zero.
answered Nov 17 at 16:13
астон вілла олоф мэллбэрг
36.7k33376
add a comment |
up vote
0
down vote
This is very simple to prove. All you need to note is the following identity :
$$
frac 1{a(a+1)...(a+m)} = frac{1}{m}left( frac 1{a(a+1)...(a+m-1)} - frac 1{(a+1)...(a+m)}right)
$$
Now, consider the sum:
$$
sum_{n=k+1}^{k+m} frac 1{binom nk} = k! times sum_{i=1}^{i=m} frac{i!}{(k+i)!} \ = k! sum_{i=1}^{i=m}frac 1{(i+1)(i+2)...(i+k)} \ = k! sum_{i=1}^{i=m} frac{1}{k-1}left(frac{1}{(i+1)...(i+k-1)} - frac 1{(i+2)...(i+k)}right) \ = frac{k!}{k-1}sum_{i=1}^{i=m} left(frac{1}{(i+1)...(i+k-1)} - frac 1{(i+2)...(i+k)}right) \ = frac{k!}{k-1} left(frac 1{k!} - frac 1{(m+2)...(m+k)}right) \ = frac 1{k-1} - frac{k!}{(k-1)(m+2)...(m+k)}
$$
By letting $m$ go to infinity, the answer is clearly $frac 1{k-1}$, since the second term goes to zero.
answered Nov 17 at 16:13
астон вілла олоф мэллбэрг
36.7k33376
add a comment |
up vote
0
down vote
up vote
0
down vote
This is very simple to prove. All you need to note is the following identity :
$$
frac 1{a(a+1)...(a+m)} = frac{1}{m}left( frac 1{a(a+1)...(a+m-1)} - frac 1{(a+1)...(a+m)}right)
$$
Now, consider the sum:
$$
sum_{n=k+1}^{k+m} frac 1{binom nk} = k! times sum_{i=1}^{i=m} frac{i!}{(k+i)!} \ = k! sum_{i=1}^{i=m}frac 1{(i+1)(i+2)...(i+k)} \ = k! sum_{i=1}^{i=m} frac{1}{k-1}left(frac{1}{(i+1)...(i+k-1)} - frac 1{(i+2)...(i+k)}right) \ = frac{k!}{k-1}sum_{i=1}^{i=m} left(frac{1}{(i+1)...(i+k-1)} - frac 1{(i+2)...(i+k)}right) \ = frac{k!}{k-1} left(frac 1{k!} - frac 1{(m+2)...(m+k)}right) \ = frac 1{k-1} - frac{k!}{(k-1)(m+2)...(m+k)}
$$
By letting $m$ go to infinity, the answer is clearly $frac 1{k-1}$, since the second term goes to zero.
answered Nov 17 at 16:13
астон вілла олоф мэллбэрг
36.7k33376
This is very simple to prove. All you need to note is the following identity :
$$
frac 1{a(a+1)...(a+m)} = frac{1}{m}left( frac 1{a(a+1)...(a+m-1)} - frac 1{(a+1)...(a+m)}right)
$$
Now, consider the sum:
$$
sum_{n=k+1}^{k+m} frac 1{binom nk} = k! times sum_{i=1}^{i=m} frac{i!}{(k+i)!} \ = k! sum_{i=1}^{i=m}frac 1{(i+1)(i+2)...(i+k)} \ = k! sum_{i=1}^{i=m} frac{1}{k-1}left(frac{1}{(i+1)...(i+k-1)} - frac 1{(i+2)...(i+k)}right) \ = frac{k!}{k-1}sum_{i=1}^{i=m} left(frac{1}{(i+1)...(i+k-1)} - frac 1{(i+2)...(i+k)}right) \ = frac{k!}{k-1} left(frac 1{k!} - frac 1{(m+2)...(m+k)}right) \ = frac 1{k-1} - frac{k!}{(k-1)(m+2)...(m+k)}
$$
By letting $m$ go to infinity, the answer is clearly $frac 1{k-1}$, since the second term goes to zero.
answered Nov 17 at 16:13
астон вілла олоф мэллбэрг
36.7k33376
answered Nov 17 at 16:13
астон вілла олоф мэллбэрг
36.7k33376
answered Nov 17 at 16:13
астон вілла олоф мэллбэрг
36.7k33376
answered Nov 17 at 16:13
астон вілла олоф мэллбэрг
36.7k33376
36.7k33376
add a comment |
add a comment |
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