Mixing
a combinatoric limit proof
$begingroup$
Let $beta >4$. Prove that if
$$a_n = sum_{k=1}^{n}k{nchoose k}{nchoose {k-1}}$$
then
$$lim_{n=0}frac{a_n}{beta^n} = 0$$
My attempt to solve it:
I tried to show that $a_n leq 4^n = sum_{k = 1}^{n}k{nchoose k}3^k$ (by derivating the binomial identity) so what I actually need to show is that ${nchoose {k-1}} leq 3^k$ for every $kin mathbb{N}$, and here I got stuck.
Any help in showing this or a completely different solution will be appreciated.
combinatorics binomial-coefficients binomial-theorem
edited Jan 13 at 20:32
leonbloy
41.1k645107
asked Jan 13 at 16:47
Robo YonuomaroRobo Yonuomaro
786
$endgroup$
add a comment |
$begingroup$
Let $beta >4$. Prove that if
$$a_n = sum_{k=1}^{n}k{nchoose k}{nchoose {k-1}}$$
then
$$lim_{n=0}frac{a_n}{beta^n} = 0$$
My attempt to solve it:
I tried to show that $a_n leq 4^n = sum_{k = 1}^{n}k{nchoose k}3^k$ (by derivating the binomial identity) so what I actually need to show is that ${nchoose {k-1}} leq 3^k$ for every $kin mathbb{N}$, and here I got stuck.
Any help in showing this or a completely different solution will be appreciated.
combinatorics binomial-coefficients binomial-theorem
edited Jan 13 at 20:32
leonbloy
41.1k645107
asked Jan 13 at 16:47
Robo YonuomaroRobo Yonuomaro
786
$endgroup$
add a comment |
$begingroup$
Let $beta >4$. Prove that if
$$a_n = sum_{k=1}^{n}k{nchoose k}{nchoose {k-1}}$$
then
$$lim_{n=0}frac{a_n}{beta^n} = 0$$
My attempt to solve it:
I tried to show that $a_n leq 4^n = sum_{k = 1}^{n}k{nchoose k}3^k$ (by derivating the binomial identity) so what I actually need to show is that ${nchoose {k-1}} leq 3^k$ for every $kin mathbb{N}$, and here I got stuck.
Any help in showing this or a completely different solution will be appreciated.
combinatorics binomial-coefficients binomial-theorem
edited Jan 13 at 20:32
leonbloy
41.1k645107
asked Jan 13 at 16:47
Robo YonuomaroRobo Yonuomaro
786
$endgroup$
Let $beta >4$. Prove that if
$$a_n = sum_{k=1}^{n}k{nchoose k}{nchoose {k-1}}$$
then
$$lim_{n=0}frac{a_n}{beta^n} = 0$$
My attempt to solve it:
I tried to show that $a_n leq 4^n = sum_{k = 1}^{n}k{nchoose k}3^k$ (by derivating the binomial identity) so what I actually need to show is that ${nchoose {k-1}} leq 3^k$ for every $kin mathbb{N}$, and here I got stuck.
Any help in showing this or a completely different solution will be appreciated.
combinatorics binomial-coefficients binomial-theorem
combinatorics binomial-coefficients binomial-theorem
edited Jan 13 at 20:32
leonbloy
41.1k645107
asked Jan 13 at 16:47
Robo YonuomaroRobo Yonuomaro
786
edited Jan 13 at 20:32
leonbloy
41.1k645107
asked Jan 13 at 16:47
Robo YonuomaroRobo Yonuomaro
786
edited Jan 13 at 20:32
leonbloy
41.1k645107
edited Jan 13 at 20:32
leonbloy
41.1k645107
edited Jan 13 at 20:32
leonbloy
41.1k645107
41.1k645107
asked Jan 13 at 16:47
Robo YonuomaroRobo Yonuomaro
786
asked Jan 13 at 16:47
Robo YonuomaroRobo Yonuomaro
786
asked Jan 13 at 16:47
Robo YonuomaroRobo Yonuomaro
786
786
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$$begin{eqnarray}a_n &=&sum_{k=0}^{n}k{nchoose k}{nchoose {k-1}}\&=&nsum_{k=0}^{n}{n-1choose k-1}{nchoose {k-1}}\
&=&nsum_{k=0}^{n}{n-1choose n-k}{nchoose {k-1}}\
&=&nbinom{2n-1}{n-1}=frac{n}{2}binom{2n}{n}.
end{eqnarray}$$ Since it holds by binomial theorem that
$$
sum_{k=0}^n binom{2n}{k}kx^{k-1}=2n(1+x)^{2n-1},
$$we have (by letting $x=1$)
$$
0le 2a_n = nbinom{2n}{n}le 2ncdot 2^{2n-1}=ncdot 4^n.
$$ For $beta>4$, we have
$$
lim_{ntoinfty} frac{a_n}{beta^n}le lim_{ntoinfty} frac{n}{2}left(frac{4}{beta}right)^n = 0.
$$
answered Jan 13 at 17:09
SongSong
13k631
$endgroup$
$begingroup$
why is $sum_{k=0}^{n}{{n-1}choose{k-n}}{nchoose {k-1}} = {{2n-1}choose{n-1}}$?
$endgroup$
– Robo Yonuomaro
Jan 13 at 17:12
$begingroup$
@RoboYonuomaro It can be obtained by comparing coefficients of $z^{n−1}$ on both sides of $$(1+z)^{n−1}(1+z)^n=(1+z)^{2n−1}.$$ And there is a typo: it should be $sum_{k=0}^{n}{{n-1}choose{n-k}}{nchoose {k-1}} =sum_{k=0}^{n-1}{{n-1}choose{n-k-1}}{nchoose {k}} = {{2n-1}choose{n-1}}.$
$endgroup$
– Song
Jan 13 at 17:29
add a comment |
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$begingroup$
$$begin{eqnarray}a_n &=&sum_{k=0}^{n}k{nchoose k}{nchoose {k-1}}\&=&nsum_{k=0}^{n}{n-1choose k-1}{nchoose {k-1}}\
&=&nsum_{k=0}^{n}{n-1choose n-k}{nchoose {k-1}}\
&=&nbinom{2n-1}{n-1}=frac{n}{2}binom{2n}{n}.
end{eqnarray}$$ Since it holds by binomial theorem that
$$
sum_{k=0}^n binom{2n}{k}kx^{k-1}=2n(1+x)^{2n-1},
$$we have (by letting $x=1$)
$$
0le 2a_n = nbinom{2n}{n}le 2ncdot 2^{2n-1}=ncdot 4^n.
$$ For $beta>4$, we have
$$
lim_{ntoinfty} frac{a_n}{beta^n}le lim_{ntoinfty} frac{n}{2}left(frac{4}{beta}right)^n = 0.
$$
answered Jan 13 at 17:09
SongSong
13k631
$endgroup$
$begingroup$
why is $sum_{k=0}^{n}{{n-1}choose{k-n}}{nchoose {k-1}} = {{2n-1}choose{n-1}}$?
$endgroup$
– Robo Yonuomaro
Jan 13 at 17:12
$begingroup$
@RoboYonuomaro It can be obtained by comparing coefficients of $z^{n−1}$ on both sides of $$(1+z)^{n−1}(1+z)^n=(1+z)^{2n−1}.$$ And there is a typo: it should be $sum_{k=0}^{n}{{n-1}choose{n-k}}{nchoose {k-1}} =sum_{k=0}^{n-1}{{n-1}choose{n-k-1}}{nchoose {k}} = {{2n-1}choose{n-1}}.$
$endgroup$
– Song
Jan 13 at 17:29
add a comment |
$begingroup$
$$begin{eqnarray}a_n &=&sum_{k=0}^{n}k{nchoose k}{nchoose {k-1}}\&=&nsum_{k=0}^{n}{n-1choose k-1}{nchoose {k-1}}\
&=&nsum_{k=0}^{n}{n-1choose n-k}{nchoose {k-1}}\
&=&nbinom{2n-1}{n-1}=frac{n}{2}binom{2n}{n}.
end{eqnarray}$$ Since it holds by binomial theorem that
$$
sum_{k=0}^n binom{2n}{k}kx^{k-1}=2n(1+x)^{2n-1},
$$we have (by letting $x=1$)
$$
0le 2a_n = nbinom{2n}{n}le 2ncdot 2^{2n-1}=ncdot 4^n.
$$ For $beta>4$, we have
$$
lim_{ntoinfty} frac{a_n}{beta^n}le lim_{ntoinfty} frac{n}{2}left(frac{4}{beta}right)^n = 0.
$$
answered Jan 13 at 17:09
SongSong
13k631
$endgroup$
$begingroup$
why is $sum_{k=0}^{n}{{n-1}choose{k-n}}{nchoose {k-1}} = {{2n-1}choose{n-1}}$?
$endgroup$
– Robo Yonuomaro
Jan 13 at 17:12
$begingroup$
@RoboYonuomaro It can be obtained by comparing coefficients of $z^{n−1}$ on both sides of $$(1+z)^{n−1}(1+z)^n=(1+z)^{2n−1}.$$ And there is a typo: it should be $sum_{k=0}^{n}{{n-1}choose{n-k}}{nchoose {k-1}} =sum_{k=0}^{n-1}{{n-1}choose{n-k-1}}{nchoose {k}} = {{2n-1}choose{n-1}}.$
$endgroup$
– Song
Jan 13 at 17:29
add a comment |
$begingroup$
$$begin{eqnarray}a_n &=&sum_{k=0}^{n}k{nchoose k}{nchoose {k-1}}\&=&nsum_{k=0}^{n}{n-1choose k-1}{nchoose {k-1}}\
&=&nsum_{k=0}^{n}{n-1choose n-k}{nchoose {k-1}}\
&=&nbinom{2n-1}{n-1}=frac{n}{2}binom{2n}{n}.
end{eqnarray}$$ Since it holds by binomial theorem that
$$
sum_{k=0}^n binom{2n}{k}kx^{k-1}=2n(1+x)^{2n-1},
$$we have (by letting $x=1$)
$$
0le 2a_n = nbinom{2n}{n}le 2ncdot 2^{2n-1}=ncdot 4^n.
$$ For $beta>4$, we have
$$
lim_{ntoinfty} frac{a_n}{beta^n}le lim_{ntoinfty} frac{n}{2}left(frac{4}{beta}right)^n = 0.
$$
answered Jan 13 at 17:09
SongSong
13k631
$endgroup$
$$begin{eqnarray}a_n &=&sum_{k=0}^{n}k{nchoose k}{nchoose {k-1}}\&=&nsum_{k=0}^{n}{n-1choose k-1}{nchoose {k-1}}\
&=&nsum_{k=0}^{n}{n-1choose n-k}{nchoose {k-1}}\
&=&nbinom{2n-1}{n-1}=frac{n}{2}binom{2n}{n}.
end{eqnarray}$$ Since it holds by binomial theorem that
$$
sum_{k=0}^n binom{2n}{k}kx^{k-1}=2n(1+x)^{2n-1},
$$we have (by letting $x=1$)
$$
0le 2a_n = nbinom{2n}{n}le 2ncdot 2^{2n-1}=ncdot 4^n.
$$ For $beta>4$, we have
$$
lim_{ntoinfty} frac{a_n}{beta^n}le lim_{ntoinfty} frac{n}{2}left(frac{4}{beta}right)^n = 0.
$$
answered Jan 13 at 17:09
SongSong
13k631
answered Jan 13 at 17:09
SongSong
13k631
answered Jan 13 at 17:09
SongSong
13k631
answered Jan 13 at 17:09
SongSong
13k631
13k631
$begingroup$
why is $sum_{k=0}^{n}{{n-1}choose{k-n}}{nchoose {k-1}} = {{2n-1}choose{n-1}}$?
$endgroup$
– Robo Yonuomaro
Jan 13 at 17:12
$begingroup$
@RoboYonuomaro It can be obtained by comparing coefficients of $z^{n−1}$ on both sides of $$(1+z)^{n−1}(1+z)^n=(1+z)^{2n−1}.$$ And there is a typo: it should be $sum_{k=0}^{n}{{n-1}choose{n-k}}{nchoose {k-1}} =sum_{k=0}^{n-1}{{n-1}choose{n-k-1}}{nchoose {k}} = {{2n-1}choose{n-1}}.$
$endgroup$
– Song
Jan 13 at 17:29
add a comment |
$begingroup$
why is $sum_{k=0}^{n}{{n-1}choose{k-n}}{nchoose {k-1}} = {{2n-1}choose{n-1}}$?
$endgroup$
– Robo Yonuomaro
Jan 13 at 17:12
$begingroup$
@RoboYonuomaro It can be obtained by comparing coefficients of $z^{n−1}$ on both sides of $$(1+z)^{n−1}(1+z)^n=(1+z)^{2n−1}.$$ And there is a typo: it should be $sum_{k=0}^{n}{{n-1}choose{n-k}}{nchoose {k-1}} =sum_{k=0}^{n-1}{{n-1}choose{n-k-1}}{nchoose {k}} = {{2n-1}choose{n-1}}.$
$endgroup$
– Song
Jan 13 at 17:29
$begingroup$
why is $sum_{k=0}^{n}{{n-1}choose{k-n}}{nchoose {k-1}} = {{2n-1}choose{n-1}}$?
$endgroup$
– Robo Yonuomaro
Jan 13 at 17:12
$begingroup$
why is $sum_{k=0}^{n}{{n-1}choose{k-n}}{nchoose {k-1}} = {{2n-1}choose{n-1}}$?
$endgroup$
– Robo Yonuomaro
Jan 13 at 17:12
$begingroup$
@RoboYonuomaro It can be obtained by comparing coefficients of $z^{n−1}$ on both sides of $$(1+z)^{n−1}(1+z)^n=(1+z)^{2n−1}.$$ And there is a typo: it should be $sum_{k=0}^{n}{{n-1}choose{n-k}}{nchoose {k-1}} =sum_{k=0}^{n-1}{{n-1}choose{n-k-1}}{nchoose {k}} = {{2n-1}choose{n-1}}.$
$endgroup$
– Song
Jan 13 at 17:29
$begingroup$
@RoboYonuomaro It can be obtained by comparing coefficients of $z^{n−1}$ on both sides of $$(1+z)^{n−1}(1+z)^n=(1+z)^{2n−1}.$$ And there is a typo: it should be $sum_{k=0}^{n}{{n-1}choose{n-k}}{nchoose {k-1}} =sum_{k=0}^{n-1}{{n-1}choose{n-k-1}}{nchoose {k}} = {{2n-1}choose{n-1}}.$
$endgroup$
– Song
Jan 13 at 17:29
add a comment |
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