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Find the sum of $2^-x/x$
Clash Royale CLAN TAG#URR8PPP
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Task is the following: find the $sum_x=1^+∞ frac2^-xx$
I don't even know how to proceed. I know that $sum_x=1^+∞ 2^-x = 1$. However, is it useful here?
summation
asked Jul 23 at 16:23
Sargis Iskandaryan
47612
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up vote
0
down vote
favorite
Task is the following: find the $sum_x=1^+∞ frac2^-xx$
I don't even know how to proceed. I know that $sum_x=1^+∞ 2^-x = 1$. However, is it useful here?
summation
asked Jul 23 at 16:23
Sargis Iskandaryan
47612
The sum is equal to $$log(2)$$
– Dr. Sonnhard Graubner
Jul 23 at 16:27
@Dr.SonnhardGraubner Yes, I know the answer. I don't know how to get it.
– Sargis Iskandaryan
Jul 23 at 16:30
1
I would consider $sum_n=1^infty 2^-n/n$ instead. The variable $x$ is usually considered to be continuous. Hint: Start with $sum_n=1^ infty x^n$, which is known. Then replace $x$ with $x/2$ to get powers of $2$. Then think about what kind of operation you can do to get the $n$ in the denominator. (Think about different things you can do in calculus...)
– Jair Taylor
Jul 23 at 16:41
1
use taylor expansion for $ln(1/2)$
– Vasya
Jul 23 at 16:41
Use this series for the natural logarithm and plug for $x= -1/2$
– Crostul
Jul 23 at 16:42
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Task is the following: find the $sum_x=1^+∞ frac2^-xx$
I don't even know how to proceed. I know that $sum_x=1^+∞ 2^-x = 1$. However, is it useful here?
summation
asked Jul 23 at 16:23
Sargis Iskandaryan
47612
Task is the following: find the $sum_x=1^+∞ frac2^-xx$
I don't even know how to proceed. I know that $sum_x=1^+∞ 2^-x = 1$. However, is it useful here?
summation
asked Jul 23 at 16:23
Sargis Iskandaryan
47612
asked Jul 23 at 16:23
Sargis Iskandaryan
47612
asked Jul 23 at 16:23
Sargis Iskandaryan
47612
asked Jul 23 at 16:23
Sargis Iskandaryan
47612
47612
The sum is equal to $$log(2)$$
– Dr. Sonnhard Graubner
Jul 23 at 16:27
@Dr.SonnhardGraubner Yes, I know the answer. I don't know how to get it.
– Sargis Iskandaryan
Jul 23 at 16:30
1
I would consider $sum_n=1^infty 2^-n/n$ instead. The variable $x$ is usually considered to be continuous. Hint: Start with $sum_n=1^ infty x^n$, which is known. Then replace $x$ with $x/2$ to get powers of $2$. Then think about what kind of operation you can do to get the $n$ in the denominator. (Think about different things you can do in calculus...)
– Jair Taylor
Jul 23 at 16:41
1
use taylor expansion for $ln(1/2)$
– Vasya
Jul 23 at 16:41
Use this series for the natural logarithm and plug for $x= -1/2$
– Crostul
Jul 23 at 16:42
add a comment |Â
The sum is equal to $$log(2)$$
– Dr. Sonnhard Graubner
Jul 23 at 16:27
@Dr.SonnhardGraubner Yes, I know the answer. I don't know how to get it.
– Sargis Iskandaryan
Jul 23 at 16:30
1
I would consider $sum_n=1^infty 2^-n/n$ instead. The variable $x$ is usually considered to be continuous. Hint: Start with $sum_n=1^ infty x^n$, which is known. Then replace $x$ with $x/2$ to get powers of $2$. Then think about what kind of operation you can do to get the $n$ in the denominator. (Think about different things you can do in calculus...)
– Jair Taylor
Jul 23 at 16:41
1
use taylor expansion for $ln(1/2)$
– Vasya
Jul 23 at 16:41
Use this series for the natural logarithm and plug for $x= -1/2$
– Crostul
Jul 23 at 16:42
The sum is equal to $$log(2)$$
– Dr. Sonnhard Graubner
Jul 23 at 16:27
The sum is equal to $$log(2)$$
– Dr. Sonnhard Graubner
Jul 23 at 16:27
@Dr.SonnhardGraubner Yes, I know the answer. I don't know how to get it.
– Sargis Iskandaryan
Jul 23 at 16:30
@Dr.SonnhardGraubner Yes, I know the answer. I don't know how to get it.
– Sargis Iskandaryan
Jul 23 at 16:30
1
1
I would consider $sum_n=1^infty 2^-n/n$ instead. The variable $x$ is usually considered to be continuous. Hint: Start with $sum_n=1^ infty x^n$, which is known. Then replace $x$ with $x/2$ to get powers of $2$. Then think about what kind of operation you can do to get the $n$ in the denominator. (Think about different things you can do in calculus...)
– Jair Taylor
Jul 23 at 16:41
I would consider $sum_n=1^infty 2^-n/n$ instead. The variable $x$ is usually considered to be continuous. Hint: Start with $sum_n=1^ infty x^n$, which is known. Then replace $x$ with $x/2$ to get powers of $2$. Then think about what kind of operation you can do to get the $n$ in the denominator. (Think about different things you can do in calculus...)
– Jair Taylor
Jul 23 at 16:41
1
1
use taylor expansion for $ln(1/2)$
– Vasya
Jul 23 at 16:41
use taylor expansion for $ln(1/2)$
– Vasya
Jul 23 at 16:41
Use this series for the natural logarithm and plug for $x= -1/2$
– Crostul
Jul 23 at 16:42
Use this series for the natural logarithm and plug for $x= -1/2$
– Crostul
Jul 23 at 16:42
add a comment |Â
3 Answers
3
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oldest
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up vote
1
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Here's a neat trick. Note that
$$ fracmathrmdmathrmdalpha sum_x=1^infty frac2^-alpha xx = -ln(2)sum_x=1^infty 2^-alpha x = -ln(2)frac2^-alpha1-2^-alpha$$
Now integrate this from $alpha=1$ to $alpha=infty$:
$$ - sum_x=1^infty frac2^-xx= left. -lnleft(1 - 2^-alpharight) right|_alpha =1^infty = -ln(2)$$
answered Jul 23 at 16:50
gj255
1,330818
add a comment |Â
up vote
3
down vote
According to Taylor's series,
$$ln(1+x)=x-fracx^22+fracx^33-fracx^44+cdots, ~~~forall x in (-1,1].$$
Let $x=-dfrac12.$ Then $$ln frac12=lnleft(1-frac12right)=-dfrac12-frac12 cdot 2^2-frac13 cdot 2^3-frac14 cdot 2^4+cdots=-sum_n=1^inftyfrac1ncdot 2^n
$$
Therefore
$$sum_n=1^inftyfrac1ncdot 2^n=-lnfrac12=ln2.$$
answered Jul 23 at 16:45
mengdie1982
2,897216
add a comment |Â
up vote
2
down vote
Hint 1: It suffices to recognize this sum as a Taylor series
Hint 2: It is in fact sufficient to know that for all $x in (-1,1)$, we have
$$
sum_x=1^infty z^x-1 = frac11-z
$$
Note that for $z = 1/2$, this is your statement. It follows that
$$
int_0^t frac11-z,dz = int_0^t sum_x=1^infty z^x-1dz =
sum_x=1^infty int_0^t z^x-1dz = sum_x=1^infty fract^xx
$$
answered Jul 23 at 16:43
Omnomnomnom
121k784170
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Here's a neat trick. Note that
$$ fracmathrmdmathrmdalpha sum_x=1^infty frac2^-alpha xx = -ln(2)sum_x=1^infty 2^-alpha x = -ln(2)frac2^-alpha1-2^-alpha$$
Now integrate this from $alpha=1$ to $alpha=infty$:
$$ - sum_x=1^infty frac2^-xx= left. -lnleft(1 - 2^-alpharight) right|_alpha =1^infty = -ln(2)$$
answered Jul 23 at 16:50
gj255
1,330818
add a comment |Â
up vote
1
down vote
accepted
Here's a neat trick. Note that
$$ fracmathrmdmathrmdalpha sum_x=1^infty frac2^-alpha xx = -ln(2)sum_x=1^infty 2^-alpha x = -ln(2)frac2^-alpha1-2^-alpha$$
Now integrate this from $alpha=1$ to $alpha=infty$:
$$ - sum_x=1^infty frac2^-xx= left. -lnleft(1 - 2^-alpharight) right|_alpha =1^infty = -ln(2)$$
answered Jul 23 at 16:50
gj255
1,330818
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Here's a neat trick. Note that
$$ fracmathrmdmathrmdalpha sum_x=1^infty frac2^-alpha xx = -ln(2)sum_x=1^infty 2^-alpha x = -ln(2)frac2^-alpha1-2^-alpha$$
Now integrate this from $alpha=1$ to $alpha=infty$:
$$ - sum_x=1^infty frac2^-xx= left. -lnleft(1 - 2^-alpharight) right|_alpha =1^infty = -ln(2)$$
answered Jul 23 at 16:50
gj255
1,330818
Here's a neat trick. Note that
$$ fracmathrmdmathrmdalpha sum_x=1^infty frac2^-alpha xx = -ln(2)sum_x=1^infty 2^-alpha x = -ln(2)frac2^-alpha1-2^-alpha$$
Now integrate this from $alpha=1$ to $alpha=infty$:
$$ - sum_x=1^infty frac2^-xx= left. -lnleft(1 - 2^-alpharight) right|_alpha =1^infty = -ln(2)$$
answered Jul 23 at 16:50
gj255
1,330818
answered Jul 23 at 16:50
gj255
1,330818
answered Jul 23 at 16:50
gj255
1,330818
answered Jul 23 at 16:50
gj255
1,330818
1,330818
add a comment |Â
add a comment |Â
up vote
3
down vote
According to Taylor's series,
$$ln(1+x)=x-fracx^22+fracx^33-fracx^44+cdots, ~~~forall x in (-1,1].$$
Let $x=-dfrac12.$ Then $$ln frac12=lnleft(1-frac12right)=-dfrac12-frac12 cdot 2^2-frac13 cdot 2^3-frac14 cdot 2^4+cdots=-sum_n=1^inftyfrac1ncdot 2^n
$$
Therefore
$$sum_n=1^inftyfrac1ncdot 2^n=-lnfrac12=ln2.$$
answered Jul 23 at 16:45
mengdie1982
2,897216
add a comment |Â
up vote
3
down vote
According to Taylor's series,
$$ln(1+x)=x-fracx^22+fracx^33-fracx^44+cdots, ~~~forall x in (-1,1].$$
Let $x=-dfrac12.$ Then $$ln frac12=lnleft(1-frac12right)=-dfrac12-frac12 cdot 2^2-frac13 cdot 2^3-frac14 cdot 2^4+cdots=-sum_n=1^inftyfrac1ncdot 2^n
$$
Therefore
$$sum_n=1^inftyfrac1ncdot 2^n=-lnfrac12=ln2.$$
answered Jul 23 at 16:45
mengdie1982
2,897216
add a comment |Â
up vote
3
down vote
up vote
3
down vote
According to Taylor's series,
$$ln(1+x)=x-fracx^22+fracx^33-fracx^44+cdots, ~~~forall x in (-1,1].$$
Let $x=-dfrac12.$ Then $$ln frac12=lnleft(1-frac12right)=-dfrac12-frac12 cdot 2^2-frac13 cdot 2^3-frac14 cdot 2^4+cdots=-sum_n=1^inftyfrac1ncdot 2^n
$$
Therefore
$$sum_n=1^inftyfrac1ncdot 2^n=-lnfrac12=ln2.$$
answered Jul 23 at 16:45
mengdie1982
2,897216
According to Taylor's series,
$$ln(1+x)=x-fracx^22+fracx^33-fracx^44+cdots, ~~~forall x in (-1,1].$$
Let $x=-dfrac12.$ Then $$ln frac12=lnleft(1-frac12right)=-dfrac12-frac12 cdot 2^2-frac13 cdot 2^3-frac14 cdot 2^4+cdots=-sum_n=1^inftyfrac1ncdot 2^n
$$
Therefore
$$sum_n=1^inftyfrac1ncdot 2^n=-lnfrac12=ln2.$$
answered Jul 23 at 16:45
mengdie1982
2,897216
answered Jul 23 at 16:45
mengdie1982
2,897216
answered Jul 23 at 16:45
mengdie1982
2,897216
answered Jul 23 at 16:45
mengdie1982
2,897216
2,897216
add a comment |Â
add a comment |Â
up vote
2
down vote
Hint 1: It suffices to recognize this sum as a Taylor series
Hint 2: It is in fact sufficient to know that for all $x in (-1,1)$, we have
$$
sum_x=1^infty z^x-1 = frac11-z
$$
Note that for $z = 1/2$, this is your statement. It follows that
$$
int_0^t frac11-z,dz = int_0^t sum_x=1^infty z^x-1dz =
sum_x=1^infty int_0^t z^x-1dz = sum_x=1^infty fract^xx
$$
answered Jul 23 at 16:43
Omnomnomnom
121k784170
add a comment |Â
up vote
2
down vote
Hint 1: It suffices to recognize this sum as a Taylor series
Hint 2: It is in fact sufficient to know that for all $x in (-1,1)$, we have
$$
sum_x=1^infty z^x-1 = frac11-z
$$
Note that for $z = 1/2$, this is your statement. It follows that
$$
int_0^t frac11-z,dz = int_0^t sum_x=1^infty z^x-1dz =
sum_x=1^infty int_0^t z^x-1dz = sum_x=1^infty fract^xx
$$
answered Jul 23 at 16:43
Omnomnomnom
121k784170
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint 1: It suffices to recognize this sum as a Taylor series
Hint 2: It is in fact sufficient to know that for all $x in (-1,1)$, we have
$$
sum_x=1^infty z^x-1 = frac11-z
$$
Note that for $z = 1/2$, this is your statement. It follows that
$$
int_0^t frac11-z,dz = int_0^t sum_x=1^infty z^x-1dz =
sum_x=1^infty int_0^t z^x-1dz = sum_x=1^infty fract^xx
$$
answered Jul 23 at 16:43
Omnomnomnom
121k784170
Hint 1: It suffices to recognize this sum as a Taylor series
Hint 2: It is in fact sufficient to know that for all $x in (-1,1)$, we have
$$
sum_x=1^infty z^x-1 = frac11-z
$$
Note that for $z = 1/2$, this is your statement. It follows that
$$
int_0^t frac11-z,dz = int_0^t sum_x=1^infty z^x-1dz =
sum_x=1^infty int_0^t z^x-1dz = sum_x=1^infty fract^xx
$$
answered Jul 23 at 16:43
Omnomnomnom
121k784170
answered Jul 23 at 16:43
Omnomnomnom
121k784170
answered Jul 23 at 16:43
Omnomnomnom
121k784170
answered Jul 23 at 16:43
Omnomnomnom
121k784170
121k784170
add a comment |Â
add a comment |Â
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The sum is equal to $$log(2)$$
– Dr. Sonnhard Graubner
Jul 23 at 16:27
@Dr.SonnhardGraubner Yes, I know the answer. I don't know how to get it.
– Sargis Iskandaryan
Jul 23 at 16:30
1
I would consider $sum_n=1^infty 2^-n/n$ instead. The variable $x$ is usually considered to be continuous. Hint: Start with $sum_n=1^ infty x^n$, which is known. Then replace $x$ with $x/2$ to get powers of $2$. Then think about what kind of operation you can do to get the $n$ in the denominator. (Think about different things you can do in calculus...)
– Jair Taylor
Jul 23 at 16:41
1
use taylor expansion for $ln(1/2)$
– Vasya
Jul 23 at 16:41
Use this series for the natural logarithm and plug for $x= -1/2$
– Crostul
Jul 23 at 16:42