Alternating sum of binomial coefficient
I'm currently playing with some sequence and series studying asymptotic properties and stuff and I'd need to know if it exists a closed form for:
$$s^k_n:=sum_{l=0}^n(-1)^l{{l+k}choose k}$$
for $kin mathbb{N}$. so far I know:
$$s^0_n=left[ frac n2 right]$$
$$s^1_n=frac {1+(2n+3)(-1)^n}4$$
any help (ways to do it by induction, recursive formula, asymptotic behavior, source, etc) is greatly appreciated.
sequences-and-series summation binomial-coefficients
asked Dec 26 at 11:27
Renato Faraone
2,32911627
add a comment |
I'm currently playing with some sequence and series studying asymptotic properties and stuff and I'd need to know if it exists a closed form for:
$$s^k_n:=sum_{l=0}^n(-1)^l{{l+k}choose k}$$
for $kin mathbb{N}$. so far I know:
$$s^0_n=left[ frac n2 right]$$
$$s^1_n=frac {1+(2n+3)(-1)^n}4$$
any help (ways to do it by induction, recursive formula, asymptotic behavior, source, etc) is greatly appreciated.
sequences-and-series summation binomial-coefficients
asked Dec 26 at 11:27
Renato Faraone
2,32911627
add a comment |
I'm currently playing with some sequence and series studying asymptotic properties and stuff and I'd need to know if it exists a closed form for:
$$s^k_n:=sum_{l=0}^n(-1)^l{{l+k}choose k}$$
for $kin mathbb{N}$. so far I know:
$$s^0_n=left[ frac n2 right]$$
$$s^1_n=frac {1+(2n+3)(-1)^n}4$$
any help (ways to do it by induction, recursive formula, asymptotic behavior, source, etc) is greatly appreciated.
sequences-and-series summation binomial-coefficients
asked Dec 26 at 11:27
Renato Faraone
2,32911627
I'm currently playing with some sequence and series studying asymptotic properties and stuff and I'd need to know if it exists a closed form for:
$$s^k_n:=sum_{l=0}^n(-1)^l{{l+k}choose k}$$
for $kin mathbb{N}$. so far I know:
$$s^0_n=left[ frac n2 right]$$
$$s^1_n=frac {1+(2n+3)(-1)^n}4$$
any help (ways to do it by induction, recursive formula, asymptotic behavior, source, etc) is greatly appreciated.
sequences-and-series summation binomial-coefficients
sequences-and-series summation binomial-coefficients
asked Dec 26 at 11:27
Renato Faraone
2,32911627
asked Dec 26 at 11:27
Renato Faraone
2,32911627
asked Dec 26 at 11:27
Renato Faraone
2,32911627
asked Dec 26 at 11:27
Renato Faraone
2,32911627
asked Dec 26 at 11:27
Renato Faraone
2,32911627
2,32911627
add a comment |
add a comment |
1 Answer
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$$s^k_n=2^{-k-1}+(-1)^nleft(^{n+k+1}_{ k}right)hypergeom([1,n+k+2],[n+2],-1)$$
$$s^k_infty=2^{-k-1}$$
answered Dec 26 at 17:56
IV_
1,118522
My guess is that it is asymptotic to $n^{k+1}$, is this correct?
– Renato Faraone
Dec 26 at 18:04
For the "sum" of the series you used Cesaro's summation right? Because that's what I'm trying to do
– Renato Faraone
3 hours ago
It's easy to prove the Cesaro sum of the alternating series is $2^{-k-1}$ I was just wondering if you used some other methods.
– Renato Faraone
1 hour ago
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$$s^k_n=2^{-k-1}+(-1)^nleft(^{n+k+1}_{ k}right)hypergeom([1,n+k+2],[n+2],-1)$$
$$s^k_infty=2^{-k-1}$$
answered Dec 26 at 17:56
IV_
1,118522
My guess is that it is asymptotic to $n^{k+1}$, is this correct?
– Renato Faraone
Dec 26 at 18:04
For the "sum" of the series you used Cesaro's summation right? Because that's what I'm trying to do
– Renato Faraone
3 hours ago
It's easy to prove the Cesaro sum of the alternating series is $2^{-k-1}$ I was just wondering if you used some other methods.
– Renato Faraone
1 hour ago
add a comment |
$$s^k_n=2^{-k-1}+(-1)^nleft(^{n+k+1}_{ k}right)hypergeom([1,n+k+2],[n+2],-1)$$
$$s^k_infty=2^{-k-1}$$
answered Dec 26 at 17:56
IV_
1,118522
My guess is that it is asymptotic to $n^{k+1}$, is this correct?
– Renato Faraone
Dec 26 at 18:04
For the "sum" of the series you used Cesaro's summation right? Because that's what I'm trying to do
– Renato Faraone
3 hours ago
It's easy to prove the Cesaro sum of the alternating series is $2^{-k-1}$ I was just wondering if you used some other methods.
– Renato Faraone
1 hour ago
add a comment |
$$s^k_n=2^{-k-1}+(-1)^nleft(^{n+k+1}_{ k}right)hypergeom([1,n+k+2],[n+2],-1)$$
$$s^k_infty=2^{-k-1}$$
answered Dec 26 at 17:56
IV_
1,118522
$$s^k_n=2^{-k-1}+(-1)^nleft(^{n+k+1}_{ k}right)hypergeom([1,n+k+2],[n+2],-1)$$
$$s^k_infty=2^{-k-1}$$
answered Dec 26 at 17:56
IV_
1,118522
edited Dec 26 at 18:24
answered Dec 26 at 17:56
IV_
1,118522
answered Dec 26 at 17:56
IV_
1,118522
answered Dec 26 at 17:56
IV_
1,118522
1,118522
My guess is that it is asymptotic to $n^{k+1}$, is this correct?
– Renato Faraone
Dec 26 at 18:04
For the "sum" of the series you used Cesaro's summation right? Because that's what I'm trying to do
– Renato Faraone
3 hours ago
It's easy to prove the Cesaro sum of the alternating series is $2^{-k-1}$ I was just wondering if you used some other methods.
– Renato Faraone
1 hour ago
add a comment |
My guess is that it is asymptotic to $n^{k+1}$, is this correct?
– Renato Faraone
Dec 26 at 18:04
For the "sum" of the series you used Cesaro's summation right? Because that's what I'm trying to do
– Renato Faraone
3 hours ago
It's easy to prove the Cesaro sum of the alternating series is $2^{-k-1}$ I was just wondering if you used some other methods.
– Renato Faraone
1 hour ago
My guess is that it is asymptotic to $n^{k+1}$, is this correct?
– Renato Faraone
Dec 26 at 18:04
My guess is that it is asymptotic to $n^{k+1}$, is this correct?
– Renato Faraone
Dec 26 at 18:04
For the "sum" of the series you used Cesaro's summation right? Because that's what I'm trying to do
– Renato Faraone
3 hours ago
For the "sum" of the series you used Cesaro's summation right? Because that's what I'm trying to do
– Renato Faraone
3 hours ago
It's easy to prove the Cesaro sum of the alternating series is $2^{-k-1}$ I was just wondering if you used some other methods.
– Renato Faraone
1 hour ago
It's easy to prove the Cesaro sum of the alternating series is $2^{-k-1}$ I was just wondering if you used some other methods.
– Renato Faraone
1 hour ago
add a comment |
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