Sum of $sum_{k=0}^{infty} k rho^{k-1}$
$begingroup$
I encounter the infinite series $sum_{k=0}^{infty} k rho^{k-1}$ in a math textbook where the answer is directly given to be $dfrac{1}{(1-rho)^2}$ when $|rho| < 1$. However, I don't understand how to obtain this result. Could someone give me a hint of how to approach this problem?
sequences-and-series analysis
asked Dec 25 '18 at 19:53
kumomkumom
82
$endgroup$
add a comment |
$begingroup$
I encounter the infinite series $sum_{k=0}^{infty} k rho^{k-1}$ in a math textbook where the answer is directly given to be $dfrac{1}{(1-rho)^2}$ when $|rho| < 1$. However, I don't understand how to obtain this result. Could someone give me a hint of how to approach this problem?
sequences-and-series analysis
asked Dec 25 '18 at 19:53
kumomkumom
82
$endgroup$
$begingroup$
See this.
$endgroup$
– David Mitra
Dec 25 '18 at 21:10
add a comment |
$begingroup$
I encounter the infinite series $sum_{k=0}^{infty} k rho^{k-1}$ in a math textbook where the answer is directly given to be $dfrac{1}{(1-rho)^2}$ when $|rho| < 1$. However, I don't understand how to obtain this result. Could someone give me a hint of how to approach this problem?
sequences-and-series analysis
asked Dec 25 '18 at 19:53
kumomkumom
82
$endgroup$
I encounter the infinite series $sum_{k=0}^{infty} k rho^{k-1}$ in a math textbook where the answer is directly given to be $dfrac{1}{(1-rho)^2}$ when $|rho| < 1$. However, I don't understand how to obtain this result. Could someone give me a hint of how to approach this problem?
sequences-and-series analysis
sequences-and-series analysis
asked Dec 25 '18 at 19:53
kumomkumom
82
asked Dec 25 '18 at 19:53
kumomkumom
82
asked Dec 25 '18 at 19:53
kumomkumom
82
asked Dec 25 '18 at 19:53
kumomkumom
82
asked Dec 25 '18 at 19:53
kumomkumom
82
82
$begingroup$
See this.
$endgroup$
– David Mitra
Dec 25 '18 at 21:10
add a comment |
$begingroup$
See this.
$endgroup$
– David Mitra
Dec 25 '18 at 21:10
$begingroup$
See this.
$endgroup$
– David Mitra
Dec 25 '18 at 21:10
$begingroup$
See this.
$endgroup$
– David Mitra
Dec 25 '18 at 21:10
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
Hint: Think of the convergent series $$sum_{k=0}^infty rho^k=frac{1}{1-rho}, qquad |rho|<1,$$ and differentiate both sides with respect to $rho$ (why is interchanging differentiation and summation possible?).
answered Dec 25 '18 at 20:02
NiklasNiklas
2,223720
$endgroup$
$begingroup$
I looked at this post and understand now why we can interchange differentiation and summation. Thanks a lot!
$endgroup$
– kumom
Dec 25 '18 at 20:29
add a comment |
$begingroup$
Start off with the geometric sum$$frac 1{1-x}=sumlimits_{ngeq0}c^n$$Now differentiate it with respect to $x$ to get$$frac 1{(1-x)^2}=sumlimits_{ngeq0}n x^{n-1}$$
answered Dec 25 '18 at 20:02
Frank W.Frank W.
3,8001321
$endgroup$
add a comment |
$begingroup$
Hint: By the geometric series, you have
$$sum_{n geq 0} r^n = frac{1}{1-r}; quad vert rvert < 1$$
Try differentiating both sides with respect to $r$ and see what you get as the result.
answered Dec 25 '18 at 20:02
KM101KM101
6,0901525
$endgroup$
add a comment |
$begingroup$
If you are familiar with the formula and derivation of the geometric series, then you can justify this by showing you can pass derivatives into a series inside the radius of convergence. This post contains a good rundown of why what we are about to do is valid.
Recall the following:
$$
sum_{k=0}^infty x^k = frac{1}{1-x} quad text{when } |x|<1
$$
If you differentiate both sides you find the following:
$$
sum_{k=0}^infty kx^{k-1} = frac{1}{(1-x)^2} quad text{when } |x|<1
$$
answered Dec 25 '18 at 20:03
ImNotTheGuyImNotTheGuy
38516
$endgroup$
add a comment |
$begingroup$
Here is a method which avoids use of derivatives or integrals. Let $S$ be the sum of the series. By distributing $rho$ among the summands of your series, show that $rho S = S - sum_{k = 0}^infty rho^k$. Use the formula for the sum of a geometric series and solve for $S$ to obtain $S = dfrac{1}{(1 - rho)^2}$.
answered Dec 25 '18 at 20:07
kobekobe
35k22248
$endgroup$
add a comment |
$begingroup$
This is an arithmetic-geometric series. One term $(k)$ is in arithmetic progression, while the other $(rho^{k-1})$ is in geometric progression. It converges for $|rho|<1$. You can find the value like this:
$displaystyle S=sum_{k=1}^{infty}krho^{k-1}=1+2rho+3rho^2+4rho^3...$
$displaystylerho S=sum_{k=1}^infty krho^k=rho+2rho^2+3rho^3+4rho^4...$
$displaystyle (1-rho)S=1+rho+rho^2...=frac1{1-rho}$ for $|rho|<1$
$displaystyletherefore S=frac1{(1-rho)^2}$
answered Dec 25 '18 at 20:07
Shubham JohriShubham Johri
5,204718
$endgroup$
add a comment |
Your Answer
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Think of the convergent series $$sum_{k=0}^infty rho^k=frac{1}{1-rho}, qquad |rho|<1,$$ and differentiate both sides with respect to $rho$ (why is interchanging differentiation and summation possible?).
answered Dec 25 '18 at 20:02
NiklasNiklas
2,223720
$endgroup$
$begingroup$
I looked at this post and understand now why we can interchange differentiation and summation. Thanks a lot!
$endgroup$
– kumom
Dec 25 '18 at 20:29
add a comment |
$begingroup$
Hint: Think of the convergent series $$sum_{k=0}^infty rho^k=frac{1}{1-rho}, qquad |rho|<1,$$ and differentiate both sides with respect to $rho$ (why is interchanging differentiation and summation possible?).
answered Dec 25 '18 at 20:02
NiklasNiklas
2,223720
$endgroup$
$begingroup$
I looked at this post and understand now why we can interchange differentiation and summation. Thanks a lot!
$endgroup$
– kumom
Dec 25 '18 at 20:29
add a comment |
$begingroup$
Hint: Think of the convergent series $$sum_{k=0}^infty rho^k=frac{1}{1-rho}, qquad |rho|<1,$$ and differentiate both sides with respect to $rho$ (why is interchanging differentiation and summation possible?).
answered Dec 25 '18 at 20:02
NiklasNiklas
2,223720
$endgroup$
Hint: Think of the convergent series $$sum_{k=0}^infty rho^k=frac{1}{1-rho}, qquad |rho|<1,$$ and differentiate both sides with respect to $rho$ (why is interchanging differentiation and summation possible?).
answered Dec 25 '18 at 20:02
NiklasNiklas
2,223720
answered Dec 25 '18 at 20:02
NiklasNiklas
2,223720
answered Dec 25 '18 at 20:02
NiklasNiklas
2,223720
answered Dec 25 '18 at 20:02
NiklasNiklas
2,223720
2,223720
$begingroup$
I looked at this post and understand now why we can interchange differentiation and summation. Thanks a lot!
$endgroup$
– kumom
Dec 25 '18 at 20:29
add a comment |
$begingroup$
I looked at this post and understand now why we can interchange differentiation and summation. Thanks a lot!
$endgroup$
– kumom
Dec 25 '18 at 20:29
$begingroup$
I looked at this post and understand now why we can interchange differentiation and summation. Thanks a lot!
$endgroup$
– kumom
Dec 25 '18 at 20:29
$begingroup$
I looked at this post and understand now why we can interchange differentiation and summation. Thanks a lot!
$endgroup$
– kumom
Dec 25 '18 at 20:29
add a comment |
$begingroup$
Start off with the geometric sum$$frac 1{1-x}=sumlimits_{ngeq0}c^n$$Now differentiate it with respect to $x$ to get$$frac 1{(1-x)^2}=sumlimits_{ngeq0}n x^{n-1}$$
answered Dec 25 '18 at 20:02
Frank W.Frank W.
3,8001321
$endgroup$
add a comment |
$begingroup$
Start off with the geometric sum$$frac 1{1-x}=sumlimits_{ngeq0}c^n$$Now differentiate it with respect to $x$ to get$$frac 1{(1-x)^2}=sumlimits_{ngeq0}n x^{n-1}$$
answered Dec 25 '18 at 20:02
Frank W.Frank W.
3,8001321
$endgroup$
add a comment |
$begingroup$
Start off with the geometric sum$$frac 1{1-x}=sumlimits_{ngeq0}c^n$$Now differentiate it with respect to $x$ to get$$frac 1{(1-x)^2}=sumlimits_{ngeq0}n x^{n-1}$$
answered Dec 25 '18 at 20:02
Frank W.Frank W.
3,8001321
$endgroup$
Start off with the geometric sum$$frac 1{1-x}=sumlimits_{ngeq0}c^n$$Now differentiate it with respect to $x$ to get$$frac 1{(1-x)^2}=sumlimits_{ngeq0}n x^{n-1}$$
answered Dec 25 '18 at 20:02
Frank W.Frank W.
3,8001321
answered Dec 25 '18 at 20:02
Frank W.Frank W.
3,8001321
answered Dec 25 '18 at 20:02
Frank W.Frank W.
3,8001321
answered Dec 25 '18 at 20:02
Frank W.Frank W.
3,8001321
3,8001321
add a comment |
add a comment |
$begingroup$
Hint: By the geometric series, you have
$$sum_{n geq 0} r^n = frac{1}{1-r}; quad vert rvert < 1$$
Try differentiating both sides with respect to $r$ and see what you get as the result.
answered Dec 25 '18 at 20:02
KM101KM101
6,0901525
$endgroup$
add a comment |
$begingroup$
Hint: By the geometric series, you have
$$sum_{n geq 0} r^n = frac{1}{1-r}; quad vert rvert < 1$$
Try differentiating both sides with respect to $r$ and see what you get as the result.
answered Dec 25 '18 at 20:02
KM101KM101
6,0901525
$endgroup$
add a comment |
$begingroup$
Hint: By the geometric series, you have
$$sum_{n geq 0} r^n = frac{1}{1-r}; quad vert rvert < 1$$
Try differentiating both sides with respect to $r$ and see what you get as the result.
answered Dec 25 '18 at 20:02
KM101KM101
6,0901525
$endgroup$
Hint: By the geometric series, you have
$$sum_{n geq 0} r^n = frac{1}{1-r}; quad vert rvert < 1$$
Try differentiating both sides with respect to $r$ and see what you get as the result.
answered Dec 25 '18 at 20:02
KM101KM101
6,0901525
answered Dec 25 '18 at 20:02
KM101KM101
6,0901525
answered Dec 25 '18 at 20:02
KM101KM101
6,0901525
answered Dec 25 '18 at 20:02
KM101KM101
6,0901525
6,0901525
add a comment |
add a comment |
$begingroup$
If you are familiar with the formula and derivation of the geometric series, then you can justify this by showing you can pass derivatives into a series inside the radius of convergence. This post contains a good rundown of why what we are about to do is valid.
Recall the following:
$$
sum_{k=0}^infty x^k = frac{1}{1-x} quad text{when } |x|<1
$$
If you differentiate both sides you find the following:
$$
sum_{k=0}^infty kx^{k-1} = frac{1}{(1-x)^2} quad text{when } |x|<1
$$
answered Dec 25 '18 at 20:03
ImNotTheGuyImNotTheGuy
38516
$endgroup$
add a comment |
$begingroup$
If you are familiar with the formula and derivation of the geometric series, then you can justify this by showing you can pass derivatives into a series inside the radius of convergence. This post contains a good rundown of why what we are about to do is valid.
Recall the following:
$$
sum_{k=0}^infty x^k = frac{1}{1-x} quad text{when } |x|<1
$$
If you differentiate both sides you find the following:
$$
sum_{k=0}^infty kx^{k-1} = frac{1}{(1-x)^2} quad text{when } |x|<1
$$
answered Dec 25 '18 at 20:03
ImNotTheGuyImNotTheGuy
38516
$endgroup$
add a comment |
$begingroup$
If you are familiar with the formula and derivation of the geometric series, then you can justify this by showing you can pass derivatives into a series inside the radius of convergence. This post contains a good rundown of why what we are about to do is valid.
Recall the following:
$$
sum_{k=0}^infty x^k = frac{1}{1-x} quad text{when } |x|<1
$$
If you differentiate both sides you find the following:
$$
sum_{k=0}^infty kx^{k-1} = frac{1}{(1-x)^2} quad text{when } |x|<1
$$
answered Dec 25 '18 at 20:03
ImNotTheGuyImNotTheGuy
38516
$endgroup$
If you are familiar with the formula and derivation of the geometric series, then you can justify this by showing you can pass derivatives into a series inside the radius of convergence. This post contains a good rundown of why what we are about to do is valid.
Recall the following:
$$
sum_{k=0}^infty x^k = frac{1}{1-x} quad text{when } |x|<1
$$
If you differentiate both sides you find the following:
$$
sum_{k=0}^infty kx^{k-1} = frac{1}{(1-x)^2} quad text{when } |x|<1
$$
answered Dec 25 '18 at 20:03
ImNotTheGuyImNotTheGuy
38516
answered Dec 25 '18 at 20:03
ImNotTheGuyImNotTheGuy
38516
answered Dec 25 '18 at 20:03
ImNotTheGuyImNotTheGuy
38516
answered Dec 25 '18 at 20:03
ImNotTheGuyImNotTheGuy
38516
38516
add a comment |
add a comment |
$begingroup$
Here is a method which avoids use of derivatives or integrals. Let $S$ be the sum of the series. By distributing $rho$ among the summands of your series, show that $rho S = S - sum_{k = 0}^infty rho^k$. Use the formula for the sum of a geometric series and solve for $S$ to obtain $S = dfrac{1}{(1 - rho)^2}$.
answered Dec 25 '18 at 20:07
kobekobe
35k22248
$endgroup$
add a comment |
$begingroup$
Here is a method which avoids use of derivatives or integrals. Let $S$ be the sum of the series. By distributing $rho$ among the summands of your series, show that $rho S = S - sum_{k = 0}^infty rho^k$. Use the formula for the sum of a geometric series and solve for $S$ to obtain $S = dfrac{1}{(1 - rho)^2}$.
answered Dec 25 '18 at 20:07
kobekobe
35k22248
$endgroup$
add a comment |
$begingroup$
Here is a method which avoids use of derivatives or integrals. Let $S$ be the sum of the series. By distributing $rho$ among the summands of your series, show that $rho S = S - sum_{k = 0}^infty rho^k$. Use the formula for the sum of a geometric series and solve for $S$ to obtain $S = dfrac{1}{(1 - rho)^2}$.
answered Dec 25 '18 at 20:07
kobekobe
35k22248
$endgroup$
Here is a method which avoids use of derivatives or integrals. Let $S$ be the sum of the series. By distributing $rho$ among the summands of your series, show that $rho S = S - sum_{k = 0}^infty rho^k$. Use the formula for the sum of a geometric series and solve for $S$ to obtain $S = dfrac{1}{(1 - rho)^2}$.
answered Dec 25 '18 at 20:07
kobekobe
35k22248
answered Dec 25 '18 at 20:07
kobekobe
35k22248
answered Dec 25 '18 at 20:07
kobekobe
35k22248
answered Dec 25 '18 at 20:07
kobekobe
35k22248
35k22248
add a comment |
add a comment |
$begingroup$
This is an arithmetic-geometric series. One term $(k)$ is in arithmetic progression, while the other $(rho^{k-1})$ is in geometric progression. It converges for $|rho|<1$. You can find the value like this:
$displaystyle S=sum_{k=1}^{infty}krho^{k-1}=1+2rho+3rho^2+4rho^3...$
$displaystylerho S=sum_{k=1}^infty krho^k=rho+2rho^2+3rho^3+4rho^4...$
$displaystyle (1-rho)S=1+rho+rho^2...=frac1{1-rho}$ for $|rho|<1$
$displaystyletherefore S=frac1{(1-rho)^2}$
answered Dec 25 '18 at 20:07
Shubham JohriShubham Johri
5,204718
$endgroup$
add a comment |
$begingroup$
This is an arithmetic-geometric series. One term $(k)$ is in arithmetic progression, while the other $(rho^{k-1})$ is in geometric progression. It converges for $|rho|<1$. You can find the value like this:
$displaystyle S=sum_{k=1}^{infty}krho^{k-1}=1+2rho+3rho^2+4rho^3...$
$displaystylerho S=sum_{k=1}^infty krho^k=rho+2rho^2+3rho^3+4rho^4...$
$displaystyle (1-rho)S=1+rho+rho^2...=frac1{1-rho}$ for $|rho|<1$
$displaystyletherefore S=frac1{(1-rho)^2}$
answered Dec 25 '18 at 20:07
Shubham JohriShubham Johri
5,204718
$endgroup$
add a comment |
$begingroup$
This is an arithmetic-geometric series. One term $(k)$ is in arithmetic progression, while the other $(rho^{k-1})$ is in geometric progression. It converges for $|rho|<1$. You can find the value like this:
$displaystyle S=sum_{k=1}^{infty}krho^{k-1}=1+2rho+3rho^2+4rho^3...$
$displaystylerho S=sum_{k=1}^infty krho^k=rho+2rho^2+3rho^3+4rho^4...$
$displaystyle (1-rho)S=1+rho+rho^2...=frac1{1-rho}$ for $|rho|<1$
$displaystyletherefore S=frac1{(1-rho)^2}$
answered Dec 25 '18 at 20:07
Shubham JohriShubham Johri
5,204718
$endgroup$
This is an arithmetic-geometric series. One term $(k)$ is in arithmetic progression, while the other $(rho^{k-1})$ is in geometric progression. It converges for $|rho|<1$. You can find the value like this:
$displaystyle S=sum_{k=1}^{infty}krho^{k-1}=1+2rho+3rho^2+4rho^3...$
$displaystylerho S=sum_{k=1}^infty krho^k=rho+2rho^2+3rho^3+4rho^4...$
$displaystyle (1-rho)S=1+rho+rho^2...=frac1{1-rho}$ for $|rho|<1$
$displaystyletherefore S=frac1{(1-rho)^2}$
answered Dec 25 '18 at 20:07
Shubham JohriShubham Johri
5,204718
answered Dec 25 '18 at 20:07
Shubham JohriShubham Johri
5,204718
answered Dec 25 '18 at 20:07
Shubham JohriShubham Johri
5,204718
answered Dec 25 '18 at 20:07
Shubham JohriShubham Johri
5,204718
5,204718
add a comment |
add a comment |
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See this.
$endgroup$
– David Mitra
Dec 25 '18 at 21:10