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Evaluating $sum_n=1^inftyfracn16^n$ [duplicate]
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Sum of a power series $n x^n$ [duplicate]
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I'm trying to evaluate the sum of the following infinite series:
$$sum_n=1^inftyfracn16^n$$
I know it converges to $frac16225$, but I don't know how to reach this solution. It's not a geometric series or a telescoping sum, and I haven't found any way to relate it to a Taylor or Maclaurin series. How should I approach this problem?
calculus sequences-and-series
asked Jul 28 at 22:22
Dibbs
154213
marked as duplicate by Hans Lundmark, Arnaud Mortier, amWhy calculus
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This question already has an answer here:
Sum of a power series $n x^n$ [duplicate]
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I'm trying to evaluate the sum of the following infinite series:
$$sum_n=1^inftyfracn16^n$$
I know it converges to $frac16225$, but I don't know how to reach this solution. It's not a geometric series or a telescoping sum, and I haven't found any way to relate it to a Taylor or Maclaurin series. How should I approach this problem?
calculus sequences-and-series
asked Jul 28 at 22:22
Dibbs
154213
marked as duplicate by Hans Lundmark, Arnaud Mortier, amWhy calculus
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up vote
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This question already has an answer here:
Sum of a power series $n x^n$ [duplicate]
2 answers
I'm trying to evaluate the sum of the following infinite series:
$$sum_n=1^inftyfracn16^n$$
I know it converges to $frac16225$, but I don't know how to reach this solution. It's not a geometric series or a telescoping sum, and I haven't found any way to relate it to a Taylor or Maclaurin series. How should I approach this problem?
calculus sequences-and-series
asked Jul 28 at 22:22
Dibbs
154213
This question already has an answer here:
Sum of a power series $n x^n$ [duplicate]
2 answers
I'm trying to evaluate the sum of the following infinite series:
$$sum_n=1^inftyfracn16^n$$
I know it converges to $frac16225$, but I don't know how to reach this solution. It's not a geometric series or a telescoping sum, and I haven't found any way to relate it to a Taylor or Maclaurin series. How should I approach this problem?
This question already has an answer here:
Sum of a power series $n x^n$ [duplicate]
2 answers
calculus sequences-and-series
asked Jul 28 at 22:22
Dibbs
154213
asked Jul 28 at 22:22
Dibbs
154213
asked Jul 28 at 22:22
Dibbs
154213
asked Jul 28 at 22:22
Dibbs
154213
154213
marked as duplicate by Hans Lundmark, Arnaud Mortier, amWhy calculus
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3 Answers
3
active
oldest
votes
up vote
6
down vote
accepted
Hint:
What is the derivative of $;sum_n=1^infty x^n$?
answered Jul 28 at 22:24
Bernard
110k635102
1
@Dibbs. When you have taken the derivative, what series do you get when you set $x=1/16$? Also, how can you rewrite $sum_n=1^infty x^n$ in a closed form (not as a series), and what is it's derivative at $x=1/16$?
– md2perpe
Jul 28 at 22:44
It took me a second to wrap my head around it, but wow, it's incredible how that works out. Thanks so much!
– Dibbs
Jul 29 at 1:01
add a comment |Â
up vote
4
down vote
Since $$sqrt[n]dfracn16^n=dfracsqrt[n]n16 to frac116<1,~~~~(n to infty)$$then by radical test, $sumlimits_n=1^inftydfracn16^n$ is convergent. Thus, we may denote
$$S=sum_n=1^inftyfracn16^n=frac116+frac216^2+frac316^3+cdots.tag1$$
Then $$16S=1+frac216+frac316^2+cdotstag2.$$
By $(2)-(1)$, we obtain $$15S=1+frac116+frac116^2+cdots=dfrac11-dfrac116=frac1615.$$
As a result, $$S=frac16225.$$
answered Jul 28 at 22:37
mengdie1982
2,827216
This helped a lot, thanks. I'm wondering, this should work for all infinite summations of the form $sum_n=1^inftyfracna^n$ where $a$ is greater than $1$, right? And the answer will always be $fraca(a-1)^2$ too, right?
– Dibbs
Jul 29 at 1:07
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up vote
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$$S=frac 1 16 +frac 2 16 ^ 2 +frac 3 16 ^ 3 +frac 4 16 ^ 4 +...\ frac S 16 =frac 1 16 ^ 2 +frac 2 16 ^ 3 +frac 3 16 ^ 4 +frac 4 16 ^ 5 +...\ S-frac S 16 =frac 1 16 +left( frac 2 16 ^ 2 -frac 1 16 ^ 2 right) +left( frac 3 16 ^ 3 -frac 2 16 ^ 3 right) +left( frac 4 16 ^ 4 -frac 3 16 ^ 4 right) +...\ frac 15 16 S=frac 1 16 +frac 1 16 ^ 2 +frac 1 16 ^ 3 +frac 1 16 ^ 4 +...\ frac 15 16 S=frac frac 1 16 1-frac 1 16 =frac 1 15 \ S=frac 16 255 \ \ $$
answered Jul 28 at 22:40
haqnatural
20.5k72457
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
Hint:
What is the derivative of $;sum_n=1^infty x^n$?
answered Jul 28 at 22:24
Bernard
110k635102
1
@Dibbs. When you have taken the derivative, what series do you get when you set $x=1/16$? Also, how can you rewrite $sum_n=1^infty x^n$ in a closed form (not as a series), and what is it's derivative at $x=1/16$?
– md2perpe
Jul 28 at 22:44
It took me a second to wrap my head around it, but wow, it's incredible how that works out. Thanks so much!
– Dibbs
Jul 29 at 1:01
add a comment |Â
up vote
6
down vote
accepted
Hint:
What is the derivative of $;sum_n=1^infty x^n$?
answered Jul 28 at 22:24
Bernard
110k635102
1
@Dibbs. When you have taken the derivative, what series do you get when you set $x=1/16$? Also, how can you rewrite $sum_n=1^infty x^n$ in a closed form (not as a series), and what is it's derivative at $x=1/16$?
– md2perpe
Jul 28 at 22:44
It took me a second to wrap my head around it, but wow, it's incredible how that works out. Thanks so much!
– Dibbs
Jul 29 at 1:01
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
Hint:
What is the derivative of $;sum_n=1^infty x^n$?
answered Jul 28 at 22:24
Bernard
110k635102
Hint:
What is the derivative of $;sum_n=1^infty x^n$?
answered Jul 28 at 22:24
Bernard
110k635102
answered Jul 28 at 22:24
Bernard
110k635102
answered Jul 28 at 22:24
Bernard
110k635102
answered Jul 28 at 22:24
Bernard
110k635102
110k635102
1
@Dibbs. When you have taken the derivative, what series do you get when you set $x=1/16$? Also, how can you rewrite $sum_n=1^infty x^n$ in a closed form (not as a series), and what is it's derivative at $x=1/16$?
– md2perpe
Jul 28 at 22:44
It took me a second to wrap my head around it, but wow, it's incredible how that works out. Thanks so much!
– Dibbs
Jul 29 at 1:01
add a comment |Â
1
@Dibbs. When you have taken the derivative, what series do you get when you set $x=1/16$? Also, how can you rewrite $sum_n=1^infty x^n$ in a closed form (not as a series), and what is it's derivative at $x=1/16$?
– md2perpe
Jul 28 at 22:44
It took me a second to wrap my head around it, but wow, it's incredible how that works out. Thanks so much!
– Dibbs
Jul 29 at 1:01
1
1
@Dibbs. When you have taken the derivative, what series do you get when you set $x=1/16$? Also, how can you rewrite $sum_n=1^infty x^n$ in a closed form (not as a series), and what is it's derivative at $x=1/16$?
– md2perpe
Jul 28 at 22:44
@Dibbs. When you have taken the derivative, what series do you get when you set $x=1/16$? Also, how can you rewrite $sum_n=1^infty x^n$ in a closed form (not as a series), and what is it's derivative at $x=1/16$?
– md2perpe
Jul 28 at 22:44
It took me a second to wrap my head around it, but wow, it's incredible how that works out. Thanks so much!
– Dibbs
Jul 29 at 1:01
It took me a second to wrap my head around it, but wow, it's incredible how that works out. Thanks so much!
– Dibbs
Jul 29 at 1:01
add a comment |Â
up vote
4
down vote
Since $$sqrt[n]dfracn16^n=dfracsqrt[n]n16 to frac116<1,~~~~(n to infty)$$then by radical test, $sumlimits_n=1^inftydfracn16^n$ is convergent. Thus, we may denote
$$S=sum_n=1^inftyfracn16^n=frac116+frac216^2+frac316^3+cdots.tag1$$
Then $$16S=1+frac216+frac316^2+cdotstag2.$$
By $(2)-(1)$, we obtain $$15S=1+frac116+frac116^2+cdots=dfrac11-dfrac116=frac1615.$$
As a result, $$S=frac16225.$$
answered Jul 28 at 22:37
mengdie1982
2,827216
This helped a lot, thanks. I'm wondering, this should work for all infinite summations of the form $sum_n=1^inftyfracna^n$ where $a$ is greater than $1$, right? And the answer will always be $fraca(a-1)^2$ too, right?
– Dibbs
Jul 29 at 1:07
add a comment |Â
up vote
4
down vote
Since $$sqrt[n]dfracn16^n=dfracsqrt[n]n16 to frac116<1,~~~~(n to infty)$$then by radical test, $sumlimits_n=1^inftydfracn16^n$ is convergent. Thus, we may denote
$$S=sum_n=1^inftyfracn16^n=frac116+frac216^2+frac316^3+cdots.tag1$$
Then $$16S=1+frac216+frac316^2+cdotstag2.$$
By $(2)-(1)$, we obtain $$15S=1+frac116+frac116^2+cdots=dfrac11-dfrac116=frac1615.$$
As a result, $$S=frac16225.$$
answered Jul 28 at 22:37
mengdie1982
2,827216
This helped a lot, thanks. I'm wondering, this should work for all infinite summations of the form $sum_n=1^inftyfracna^n$ where $a$ is greater than $1$, right? And the answer will always be $fraca(a-1)^2$ too, right?
– Dibbs
Jul 29 at 1:07
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Since $$sqrt[n]dfracn16^n=dfracsqrt[n]n16 to frac116<1,~~~~(n to infty)$$then by radical test, $sumlimits_n=1^inftydfracn16^n$ is convergent. Thus, we may denote
$$S=sum_n=1^inftyfracn16^n=frac116+frac216^2+frac316^3+cdots.tag1$$
Then $$16S=1+frac216+frac316^2+cdotstag2.$$
By $(2)-(1)$, we obtain $$15S=1+frac116+frac116^2+cdots=dfrac11-dfrac116=frac1615.$$
As a result, $$S=frac16225.$$
answered Jul 28 at 22:37
mengdie1982
2,827216
Since $$sqrt[n]dfracn16^n=dfracsqrt[n]n16 to frac116<1,~~~~(n to infty)$$then by radical test, $sumlimits_n=1^inftydfracn16^n$ is convergent. Thus, we may denote
$$S=sum_n=1^inftyfracn16^n=frac116+frac216^2+frac316^3+cdots.tag1$$
Then $$16S=1+frac216+frac316^2+cdotstag2.$$
By $(2)-(1)$, we obtain $$15S=1+frac116+frac116^2+cdots=dfrac11-dfrac116=frac1615.$$
As a result, $$S=frac16225.$$
answered Jul 28 at 22:37
mengdie1982
2,827216
answered Jul 28 at 22:37
mengdie1982
2,827216
answered Jul 28 at 22:37
mengdie1982
2,827216
answered Jul 28 at 22:37
mengdie1982
2,827216
2,827216
This helped a lot, thanks. I'm wondering, this should work for all infinite summations of the form $sum_n=1^inftyfracna^n$ where $a$ is greater than $1$, right? And the answer will always be $fraca(a-1)^2$ too, right?
– Dibbs
Jul 29 at 1:07
add a comment |Â
This helped a lot, thanks. I'm wondering, this should work for all infinite summations of the form $sum_n=1^inftyfracna^n$ where $a$ is greater than $1$, right? And the answer will always be $fraca(a-1)^2$ too, right?
– Dibbs
Jul 29 at 1:07
This helped a lot, thanks. I'm wondering, this should work for all infinite summations of the form $sum_n=1^inftyfracna^n$ where $a$ is greater than $1$, right? And the answer will always be $fraca(a-1)^2$ too, right?
– Dibbs
Jul 29 at 1:07
This helped a lot, thanks. I'm wondering, this should work for all infinite summations of the form $sum_n=1^inftyfracna^n$ where $a$ is greater than $1$, right? And the answer will always be $fraca(a-1)^2$ too, right?
– Dibbs
Jul 29 at 1:07
add a comment |Â
up vote
3
down vote
$$S=frac 1 16 +frac 2 16 ^ 2 +frac 3 16 ^ 3 +frac 4 16 ^ 4 +...\ frac S 16 =frac 1 16 ^ 2 +frac 2 16 ^ 3 +frac 3 16 ^ 4 +frac 4 16 ^ 5 +...\ S-frac S 16 =frac 1 16 +left( frac 2 16 ^ 2 -frac 1 16 ^ 2 right) +left( frac 3 16 ^ 3 -frac 2 16 ^ 3 right) +left( frac 4 16 ^ 4 -frac 3 16 ^ 4 right) +...\ frac 15 16 S=frac 1 16 +frac 1 16 ^ 2 +frac 1 16 ^ 3 +frac 1 16 ^ 4 +...\ frac 15 16 S=frac frac 1 16 1-frac 1 16 =frac 1 15 \ S=frac 16 255 \ \ $$
answered Jul 28 at 22:40
haqnatural
20.5k72457
add a comment |Â
up vote
3
down vote
$$S=frac 1 16 +frac 2 16 ^ 2 +frac 3 16 ^ 3 +frac 4 16 ^ 4 +...\ frac S 16 =frac 1 16 ^ 2 +frac 2 16 ^ 3 +frac 3 16 ^ 4 +frac 4 16 ^ 5 +...\ S-frac S 16 =frac 1 16 +left( frac 2 16 ^ 2 -frac 1 16 ^ 2 right) +left( frac 3 16 ^ 3 -frac 2 16 ^ 3 right) +left( frac 4 16 ^ 4 -frac 3 16 ^ 4 right) +...\ frac 15 16 S=frac 1 16 +frac 1 16 ^ 2 +frac 1 16 ^ 3 +frac 1 16 ^ 4 +...\ frac 15 16 S=frac frac 1 16 1-frac 1 16 =frac 1 15 \ S=frac 16 255 \ \ $$
answered Jul 28 at 22:40
haqnatural
20.5k72457
add a comment |Â
up vote
3
down vote
up vote
3
down vote
$$S=frac 1 16 +frac 2 16 ^ 2 +frac 3 16 ^ 3 +frac 4 16 ^ 4 +...\ frac S 16 =frac 1 16 ^ 2 +frac 2 16 ^ 3 +frac 3 16 ^ 4 +frac 4 16 ^ 5 +...\ S-frac S 16 =frac 1 16 +left( frac 2 16 ^ 2 -frac 1 16 ^ 2 right) +left( frac 3 16 ^ 3 -frac 2 16 ^ 3 right) +left( frac 4 16 ^ 4 -frac 3 16 ^ 4 right) +...\ frac 15 16 S=frac 1 16 +frac 1 16 ^ 2 +frac 1 16 ^ 3 +frac 1 16 ^ 4 +...\ frac 15 16 S=frac frac 1 16 1-frac 1 16 =frac 1 15 \ S=frac 16 255 \ \ $$
answered Jul 28 at 22:40
haqnatural
20.5k72457
$$S=frac 1 16 +frac 2 16 ^ 2 +frac 3 16 ^ 3 +frac 4 16 ^ 4 +...\ frac S 16 =frac 1 16 ^ 2 +frac 2 16 ^ 3 +frac 3 16 ^ 4 +frac 4 16 ^ 5 +...\ S-frac S 16 =frac 1 16 +left( frac 2 16 ^ 2 -frac 1 16 ^ 2 right) +left( frac 3 16 ^ 3 -frac 2 16 ^ 3 right) +left( frac 4 16 ^ 4 -frac 3 16 ^ 4 right) +...\ frac 15 16 S=frac 1 16 +frac 1 16 ^ 2 +frac 1 16 ^ 3 +frac 1 16 ^ 4 +...\ frac 15 16 S=frac frac 1 16 1-frac 1 16 =frac 1 15 \ S=frac 16 255 \ \ $$
answered Jul 28 at 22:40
haqnatural
20.5k72457
answered Jul 28 at 22:40
haqnatural
20.5k72457
answered Jul 28 at 22:40
haqnatural
20.5k72457
answered Jul 28 at 22:40
haqnatural
20.5k72457
20.5k72457
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