Using a power series to show $log(2) = sum_{k = 1}^{infty} frac{1}{k cdot 2^{k}}$ [duplicate]
This question already has an answer here:
Show $ln2 = sum_limits{n=1}^inftyfrac1{n2^n}$
4 answers
I want to show
$$log(2) = sum_{k = 1}^{infty} frac{1}{k cdot 2^{k}}.$$
So, I start with the geometric series:
$$frac{1}{1 - x} = sum_{n = 0}^{infty} x^{n} $$
Then take the integral to get
$$log(1 - x) = -sum_{n = 1}^{infty} frac{x^{n}}{n}.$$
Then plug in $x = -1$ to get
$$log(2) = -sum_{n = 1}^{infty} frac{(-1)^{n}}{n} = sum_{n = 1}^{infty} frac{(-1)^{n + 1}}{n},$$
but this isn't equal to the original equality. Why not? How can I do the problem?
sequences-and-series power-series taylor-expansion
asked Nov 26 at 23:42
stackofhay42
1696
marked as duplicate by Did sequences-and-series
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Nov 27 at 0:31
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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This question already has an answer here:
Show $ln2 = sum_limits{n=1}^inftyfrac1{n2^n}$
4 answers
I want to show
$$log(2) = sum_{k = 1}^{infty} frac{1}{k cdot 2^{k}}.$$
So, I start with the geometric series:
$$frac{1}{1 - x} = sum_{n = 0}^{infty} x^{n} $$
Then take the integral to get
$$log(1 - x) = -sum_{n = 1}^{infty} frac{x^{n}}{n}.$$
Then plug in $x = -1$ to get
$$log(2) = -sum_{n = 1}^{infty} frac{(-1)^{n}}{n} = sum_{n = 1}^{infty} frac{(-1)^{n + 1}}{n},$$
but this isn't equal to the original equality. Why not? How can I do the problem?
sequences-and-series power-series taylor-expansion
asked Nov 26 at 23:42
stackofhay42
1696
marked as duplicate by Did sequences-and-series
Users with the sequences-and-series badge can single-handedly close sequences-and-series questions as duplicates and reopen them as needed.
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Nov 27 at 0:31
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
Show $ln2 = sum_limits{n=1}^inftyfrac1{n2^n}$
4 answers
I want to show
$$log(2) = sum_{k = 1}^{infty} frac{1}{k cdot 2^{k}}.$$
So, I start with the geometric series:
$$frac{1}{1 - x} = sum_{n = 0}^{infty} x^{n} $$
Then take the integral to get
$$log(1 - x) = -sum_{n = 1}^{infty} frac{x^{n}}{n}.$$
Then plug in $x = -1$ to get
$$log(2) = -sum_{n = 1}^{infty} frac{(-1)^{n}}{n} = sum_{n = 1}^{infty} frac{(-1)^{n + 1}}{n},$$
but this isn't equal to the original equality. Why not? How can I do the problem?
sequences-and-series power-series taylor-expansion
asked Nov 26 at 23:42
stackofhay42
1696
This question already has an answer here:
Show $ln2 = sum_limits{n=1}^inftyfrac1{n2^n}$
4 answers
I want to show
$$log(2) = sum_{k = 1}^{infty} frac{1}{k cdot 2^{k}}.$$
So, I start with the geometric series:
$$frac{1}{1 - x} = sum_{n = 0}^{infty} x^{n} $$
Then take the integral to get
$$log(1 - x) = -sum_{n = 1}^{infty} frac{x^{n}}{n}.$$
Then plug in $x = -1$ to get
$$log(2) = -sum_{n = 1}^{infty} frac{(-1)^{n}}{n} = sum_{n = 1}^{infty} frac{(-1)^{n + 1}}{n},$$
but this isn't equal to the original equality. Why not? How can I do the problem?
This question already has an answer here:
Show $ln2 = sum_limits{n=1}^inftyfrac1{n2^n}$
4 answers
sequences-and-series power-series taylor-expansion
sequences-and-series power-series taylor-expansion
asked Nov 26 at 23:42
stackofhay42
1696
asked Nov 26 at 23:42
stackofhay42
1696
asked Nov 26 at 23:42
stackofhay42
1696
asked Nov 26 at 23:42
stackofhay42
1696
asked Nov 26 at 23:42
stackofhay42
1696
1696
marked as duplicate by Did sequences-and-series
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Nov 27 at 0:31
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Did sequences-and-series
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Nov 27 at 0:31
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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2 Answers
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Plug in $x=dfrac12$ instead, and you will get the equality that you're after (since $logleft(frac12right)=-log(2)$).
answered Nov 26 at 23:45
José Carlos Santos
149k22117219
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Since the series are convergent for $|x| < 1$ then the "trick" is to use $x=1/2$. This leads to
begin{align}
frac{1}{1 - x} &= sum_{n = 0}^{infty} x^{n} \
log(1 - x) &= -sum_{n = 1}^{infty} frac{x^{n}}{n}
end{align}
becoming
begin{align}
2 &= sum_{n=0}^{infty} frac{1}{2^{n}} \
- lnleft(1 - frac{1}{2}right) = ln(2) &= sum_{n=1}^{infty} frac{1}{2^{n} , n}
end{align}
answered Nov 26 at 23:49
Leucippus
19.6k102871
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Plug in $x=dfrac12$ instead, and you will get the equality that you're after (since $logleft(frac12right)=-log(2)$).
answered Nov 26 at 23:45
José Carlos Santos
149k22117219
add a comment |
Plug in $x=dfrac12$ instead, and you will get the equality that you're after (since $logleft(frac12right)=-log(2)$).
answered Nov 26 at 23:45
José Carlos Santos
149k22117219
add a comment |
Plug in $x=dfrac12$ instead, and you will get the equality that you're after (since $logleft(frac12right)=-log(2)$).
answered Nov 26 at 23:45
José Carlos Santos
149k22117219
Plug in $x=dfrac12$ instead, and you will get the equality that you're after (since $logleft(frac12right)=-log(2)$).
answered Nov 26 at 23:45
José Carlos Santos
149k22117219
answered Nov 26 at 23:45
José Carlos Santos
149k22117219
answered Nov 26 at 23:45
José Carlos Santos
149k22117219
answered Nov 26 at 23:45
José Carlos Santos
149k22117219
149k22117219
add a comment |
add a comment |
Since the series are convergent for $|x| < 1$ then the "trick" is to use $x=1/2$. This leads to
begin{align}
frac{1}{1 - x} &= sum_{n = 0}^{infty} x^{n} \
log(1 - x) &= -sum_{n = 1}^{infty} frac{x^{n}}{n}
end{align}
becoming
begin{align}
2 &= sum_{n=0}^{infty} frac{1}{2^{n}} \
- lnleft(1 - frac{1}{2}right) = ln(2) &= sum_{n=1}^{infty} frac{1}{2^{n} , n}
end{align}
answered Nov 26 at 23:49
Leucippus
19.6k102871
add a comment |
Since the series are convergent for $|x| < 1$ then the "trick" is to use $x=1/2$. This leads to
begin{align}
frac{1}{1 - x} &= sum_{n = 0}^{infty} x^{n} \
log(1 - x) &= -sum_{n = 1}^{infty} frac{x^{n}}{n}
end{align}
becoming
begin{align}
2 &= sum_{n=0}^{infty} frac{1}{2^{n}} \
- lnleft(1 - frac{1}{2}right) = ln(2) &= sum_{n=1}^{infty} frac{1}{2^{n} , n}
end{align}
answered Nov 26 at 23:49
Leucippus
19.6k102871
add a comment |
Since the series are convergent for $|x| < 1$ then the "trick" is to use $x=1/2$. This leads to
begin{align}
frac{1}{1 - x} &= sum_{n = 0}^{infty} x^{n} \
log(1 - x) &= -sum_{n = 1}^{infty} frac{x^{n}}{n}
end{align}
becoming
begin{align}
2 &= sum_{n=0}^{infty} frac{1}{2^{n}} \
- lnleft(1 - frac{1}{2}right) = ln(2) &= sum_{n=1}^{infty} frac{1}{2^{n} , n}
end{align}
answered Nov 26 at 23:49
Leucippus
19.6k102871
Since the series are convergent for $|x| < 1$ then the "trick" is to use $x=1/2$. This leads to
begin{align}
frac{1}{1 - x} &= sum_{n = 0}^{infty} x^{n} \
log(1 - x) &= -sum_{n = 1}^{infty} frac{x^{n}}{n}
end{align}
becoming
begin{align}
2 &= sum_{n=0}^{infty} frac{1}{2^{n}} \
- lnleft(1 - frac{1}{2}right) = ln(2) &= sum_{n=1}^{infty} frac{1}{2^{n} , n}
end{align}
answered Nov 26 at 23:49
Leucippus
19.6k102871
answered Nov 26 at 23:49
Leucippus
19.6k102871
answered Nov 26 at 23:49
Leucippus
19.6k102871
answered Nov 26 at 23:49
Leucippus
19.6k102871
19.6k102871
add a comment |
add a comment |
