Mixing
$sum_{k=1}^{infty} k {n+k choose n} p^k$
$begingroup$
How can I possibly calculate
$$sum_{k=1}^{infty} k {n+k choose n} p^k$$ ???$$
I tried naming that as $U_n$ and used the fact that the sum is infinite and that
$$^{n}C_p = ^{n-1}C_p + ^{n-1}C_{p-1}$$
But I arrive at $$U_n = pU_n + U_{n-1} + T_n$$
Where $$T_n = sum_{k=1}^{infty} {{n+k-1} choose n} p^k$$ another sum I can't calculate
So I think my approach is wrong
Any help would be appreciated
Thanks in advance!!
sequences-and-series combinations
edited Jan 3 at 8:50
Arjang
5,58562363
asked Jan 3 at 0:38
arsene steinarsene stein
1096
$endgroup$
add a comment |
$begingroup$
How can I possibly calculate
$$sum_{k=1}^{infty} k {n+k choose n} p^k$$ ???$$
I tried naming that as $U_n$ and used the fact that the sum is infinite and that
$$^{n}C_p = ^{n-1}C_p + ^{n-1}C_{p-1}$$
But I arrive at $$U_n = pU_n + U_{n-1} + T_n$$
Where $$T_n = sum_{k=1}^{infty} {{n+k-1} choose n} p^k$$ another sum I can't calculate
So I think my approach is wrong
Any help would be appreciated
Thanks in advance!!
sequences-and-series combinations
edited Jan 3 at 8:50
Arjang
5,58562363
asked Jan 3 at 0:38
arsene steinarsene stein
1096
$endgroup$
$begingroup$
Do you mean $sum_{k=1}^infty k {{n+k} choose n} p^k$?
$endgroup$
– Toby Mak
Jan 3 at 0:42
$begingroup$
Yes that's it thanks for the correction
$endgroup$
– arsene stein
Jan 3 at 0:44
add a comment |
$begingroup$
How can I possibly calculate
$$sum_{k=1}^{infty} k {n+k choose n} p^k$$ ???$$
I tried naming that as $U_n$ and used the fact that the sum is infinite and that
$$^{n}C_p = ^{n-1}C_p + ^{n-1}C_{p-1}$$
But I arrive at $$U_n = pU_n + U_{n-1} + T_n$$
Where $$T_n = sum_{k=1}^{infty} {{n+k-1} choose n} p^k$$ another sum I can't calculate
So I think my approach is wrong
Any help would be appreciated
Thanks in advance!!
sequences-and-series combinations
edited Jan 3 at 8:50
Arjang
5,58562363
asked Jan 3 at 0:38
arsene steinarsene stein
1096
$endgroup$
How can I possibly calculate
$$sum_{k=1}^{infty} k {n+k choose n} p^k$$ ???$$
I tried naming that as $U_n$ and used the fact that the sum is infinite and that
$$^{n}C_p = ^{n-1}C_p + ^{n-1}C_{p-1}$$
But I arrive at $$U_n = pU_n + U_{n-1} + T_n$$
Where $$T_n = sum_{k=1}^{infty} {{n+k-1} choose n} p^k$$ another sum I can't calculate
So I think my approach is wrong
Any help would be appreciated
Thanks in advance!!
sequences-and-series combinations
sequences-and-series combinations
edited Jan 3 at 8:50
Arjang
5,58562363
asked Jan 3 at 0:38
arsene steinarsene stein
1096
edited Jan 3 at 8:50
Arjang
5,58562363
asked Jan 3 at 0:38
arsene steinarsene stein
1096
edited Jan 3 at 8:50
Arjang
5,58562363
edited Jan 3 at 8:50
Arjang
5,58562363
edited Jan 3 at 8:50
Arjang
5,58562363
5,58562363
asked Jan 3 at 0:38
arsene steinarsene stein
1096
asked Jan 3 at 0:38
arsene steinarsene stein
1096
asked Jan 3 at 0:38
arsene steinarsene stein
1096
1096
$begingroup$
Do you mean $sum_{k=1}^infty k {{n+k} choose n} p^k$?
$endgroup$
– Toby Mak
Jan 3 at 0:42
$begingroup$
Yes that's it thanks for the correction
$endgroup$
– arsene stein
Jan 3 at 0:44
add a comment |
$begingroup$
Do you mean $sum_{k=1}^infty k {{n+k} choose n} p^k$?
$endgroup$
– Toby Mak
Jan 3 at 0:42
$begingroup$
Yes that's it thanks for the correction
$endgroup$
– arsene stein
Jan 3 at 0:44
$begingroup$
Do you mean $sum_{k=1}^infty k {{n+k} choose n} p^k$?
$endgroup$
– Toby Mak
Jan 3 at 0:42
$begingroup$
Do you mean $sum_{k=1}^infty k {{n+k} choose n} p^k$?
$endgroup$
– Toby Mak
Jan 3 at 0:42
$begingroup$
Yes that's it thanks for the correction
$endgroup$
– arsene stein
Jan 3 at 0:44
$begingroup$
Yes that's it thanks for the correction
$endgroup$
– arsene stein
Jan 3 at 0:44
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
It makes our exercise easier if we use Newton's binomial series (see https://en.wikipedia.org/wiki/Binomial_coefficient)
First, convert the sum as follows:
$sumlimits_{k=0}^{infty} k {n+k choose k} p^k=p frac{d}{dp}sumlimits_{k=0}^n p^k {n+k choose k} $
Accordance with the alternative expression of Newton's binomial series we know that
$sumlimits_{k=0}^n p^k {n+k choose k}=frac{1}{(1-p)^{n+1}} $
After derivation we get the result:
$sumlimits_{k=1}^{infty} k {n+k choose n} p^k=frac {p(n+1)}{(1-p)^{(n+2)}}$
answered Jan 3 at 8:42
JV.StalkerJV.Stalker
57839
$endgroup$
$begingroup$
Thanks so much it's clearer now
$endgroup$
– arsene stein
Jan 3 at 9:38
$begingroup$
You are welcome!
$endgroup$
– JV.Stalker
Jan 3 at 9:55
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It makes our exercise easier if we use Newton's binomial series (see https://en.wikipedia.org/wiki/Binomial_coefficient)
First, convert the sum as follows:
$sumlimits_{k=0}^{infty} k {n+k choose k} p^k=p frac{d}{dp}sumlimits_{k=0}^n p^k {n+k choose k} $
Accordance with the alternative expression of Newton's binomial series we know that
$sumlimits_{k=0}^n p^k {n+k choose k}=frac{1}{(1-p)^{n+1}} $
After derivation we get the result:
$sumlimits_{k=1}^{infty} k {n+k choose n} p^k=frac {p(n+1)}{(1-p)^{(n+2)}}$
answered Jan 3 at 8:42
JV.StalkerJV.Stalker
57839
$endgroup$
$begingroup$
Thanks so much it's clearer now
$endgroup$
– arsene stein
Jan 3 at 9:38
$begingroup$
You are welcome!
$endgroup$
– JV.Stalker
Jan 3 at 9:55
add a comment |
$begingroup$
It makes our exercise easier if we use Newton's binomial series (see https://en.wikipedia.org/wiki/Binomial_coefficient)
First, convert the sum as follows:
$sumlimits_{k=0}^{infty} k {n+k choose k} p^k=p frac{d}{dp}sumlimits_{k=0}^n p^k {n+k choose k} $
Accordance with the alternative expression of Newton's binomial series we know that
$sumlimits_{k=0}^n p^k {n+k choose k}=frac{1}{(1-p)^{n+1}} $
After derivation we get the result:
$sumlimits_{k=1}^{infty} k {n+k choose n} p^k=frac {p(n+1)}{(1-p)^{(n+2)}}$
answered Jan 3 at 8:42
JV.StalkerJV.Stalker
57839
$endgroup$
$begingroup$
Thanks so much it's clearer now
$endgroup$
– arsene stein
Jan 3 at 9:38
$begingroup$
You are welcome!
$endgroup$
– JV.Stalker
Jan 3 at 9:55
add a comment |
$begingroup$
It makes our exercise easier if we use Newton's binomial series (see https://en.wikipedia.org/wiki/Binomial_coefficient)
First, convert the sum as follows:
$sumlimits_{k=0}^{infty} k {n+k choose k} p^k=p frac{d}{dp}sumlimits_{k=0}^n p^k {n+k choose k} $
Accordance with the alternative expression of Newton's binomial series we know that
$sumlimits_{k=0}^n p^k {n+k choose k}=frac{1}{(1-p)^{n+1}} $
After derivation we get the result:
$sumlimits_{k=1}^{infty} k {n+k choose n} p^k=frac {p(n+1)}{(1-p)^{(n+2)}}$
answered Jan 3 at 8:42
JV.StalkerJV.Stalker
57839
$endgroup$
It makes our exercise easier if we use Newton's binomial series (see https://en.wikipedia.org/wiki/Binomial_coefficient)
First, convert the sum as follows:
$sumlimits_{k=0}^{infty} k {n+k choose k} p^k=p frac{d}{dp}sumlimits_{k=0}^n p^k {n+k choose k} $
Accordance with the alternative expression of Newton's binomial series we know that
$sumlimits_{k=0}^n p^k {n+k choose k}=frac{1}{(1-p)^{n+1}} $
After derivation we get the result:
$sumlimits_{k=1}^{infty} k {n+k choose n} p^k=frac {p(n+1)}{(1-p)^{(n+2)}}$
answered Jan 3 at 8:42
JV.StalkerJV.Stalker
57839
answered Jan 3 at 8:42
JV.StalkerJV.Stalker
57839
answered Jan 3 at 8:42
JV.StalkerJV.Stalker
57839
answered Jan 3 at 8:42
JV.StalkerJV.Stalker
57839
57839
$begingroup$
Thanks so much it's clearer now
$endgroup$
– arsene stein
Jan 3 at 9:38
$begingroup$
You are welcome!
$endgroup$
– JV.Stalker
Jan 3 at 9:55
add a comment |
$begingroup$
Thanks so much it's clearer now
$endgroup$
– arsene stein
Jan 3 at 9:38
$begingroup$
You are welcome!
$endgroup$
– JV.Stalker
Jan 3 at 9:55
$begingroup$
Thanks so much it's clearer now
$endgroup$
– arsene stein
Jan 3 at 9:38
$begingroup$
Thanks so much it's clearer now
$endgroup$
– arsene stein
Jan 3 at 9:38
$begingroup$
You are welcome!
$endgroup$
– JV.Stalker
Jan 3 at 9:55
$begingroup$
You are welcome!
$endgroup$
– JV.Stalker
Jan 3 at 9:55
add a comment |
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$begingroup$
Do you mean $sum_{k=1}^infty k {{n+k} choose n} p^k$?
$endgroup$
– Toby Mak
Jan 3 at 0:42
$begingroup$
Yes that's it thanks for the correction
$endgroup$
– arsene stein
Jan 3 at 0:44