Why does $ 1 + frac 13 . frac 14 + frac 15 . frac 1{4^2} + frac 17 . frac 1{4^3} + … = log 3 $? [duplicate]
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This question already has an answer here:
Evaluate $ 1 + frac{1}{3}frac{1}{4}+frac{1}{5}frac{1}{4^2}+frac{1}{7}frac{1}{4^3}+dots$
1 answer
$ 1 + frac 13 . frac 14 + frac 15 . frac 1{4^2} + frac 17 . frac 1{4^3} + ...$
I evaluated the expression for the first few terms and I find that this number will probably tend to $ log 3 $. I'd like to know why, or how I can prove that it does indeed tend to $ log 3 $. More importantly, I'd like to know the relationship between this particular type of a series and the natural logarithm of numbers.
Why does the natural logarithm of a number show up in such a series?
sequences-and-series logarithms
edited Jan 3 at 18:17
Eevee Trainer
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asked Jan 3 at 17:53
WorldGovWorldGov
345211
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marked as duplicate by Martin R, Did, José Carlos Santos sequences-and-series
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Jan 3 at 17:59
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 |
$begingroup$
This question already has an answer here:
Evaluate $ 1 + frac{1}{3}frac{1}{4}+frac{1}{5}frac{1}{4^2}+frac{1}{7}frac{1}{4^3}+dots$
1 answer
$ 1 + frac 13 . frac 14 + frac 15 . frac 1{4^2} + frac 17 . frac 1{4^3} + ...$
I evaluated the expression for the first few terms and I find that this number will probably tend to $ log 3 $. I'd like to know why, or how I can prove that it does indeed tend to $ log 3 $. More importantly, I'd like to know the relationship between this particular type of a series and the natural logarithm of numbers.
Why does the natural logarithm of a number show up in such a series?
sequences-and-series logarithms
edited Jan 3 at 18:17
Eevee Trainer
9,39331640
asked Jan 3 at 17:53
WorldGovWorldGov
345211
$endgroup$
marked as duplicate by Martin R, Did, José Carlos Santos sequences-and-series
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Jan 3 at 17:59
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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Looks like: $$sum_{r=0}^{infty}{frac{1}{4^r(2r+1)}}$$
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– Rhys Hughes
Jan 3 at 17:59
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The general term is $[(2n+1)4^n]^{-1}$ if that helps.
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– John
Jan 3 at 17:59
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The reason $log$ appears is simple: its derivative is $frac1x$, which is exactly what allows you to get all the odd numbers to appear in the denominators..
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– Arthur
Jan 3 at 18:00
3
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You should be careful about writing $+dotsinfty$, it looks like you're adding infinity to all the other fractions. This, however, is not what you mean.
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– Michael Burr
Jan 3 at 18:02
add a comment |
$begingroup$
This question already has an answer here:
Evaluate $ 1 + frac{1}{3}frac{1}{4}+frac{1}{5}frac{1}{4^2}+frac{1}{7}frac{1}{4^3}+dots$
1 answer
$ 1 + frac 13 . frac 14 + frac 15 . frac 1{4^2} + frac 17 . frac 1{4^3} + ...$
I evaluated the expression for the first few terms and I find that this number will probably tend to $ log 3 $. I'd like to know why, or how I can prove that it does indeed tend to $ log 3 $. More importantly, I'd like to know the relationship between this particular type of a series and the natural logarithm of numbers.
Why does the natural logarithm of a number show up in such a series?
sequences-and-series logarithms
edited Jan 3 at 18:17
Eevee Trainer
9,39331640
asked Jan 3 at 17:53
WorldGovWorldGov
345211
$endgroup$
This question already has an answer here:
Evaluate $ 1 + frac{1}{3}frac{1}{4}+frac{1}{5}frac{1}{4^2}+frac{1}{7}frac{1}{4^3}+dots$
1 answer
$ 1 + frac 13 . frac 14 + frac 15 . frac 1{4^2} + frac 17 . frac 1{4^3} + ...$
I evaluated the expression for the first few terms and I find that this number will probably tend to $ log 3 $. I'd like to know why, or how I can prove that it does indeed tend to $ log 3 $. More importantly, I'd like to know the relationship between this particular type of a series and the natural logarithm of numbers.
Why does the natural logarithm of a number show up in such a series?
This question already has an answer here:
Evaluate $ 1 + frac{1}{3}frac{1}{4}+frac{1}{5}frac{1}{4^2}+frac{1}{7}frac{1}{4^3}+dots$
1 answer
sequences-and-series logarithms
sequences-and-series logarithms
edited Jan 3 at 18:17
Eevee Trainer
9,39331640
asked Jan 3 at 17:53
WorldGovWorldGov
345211
edited Jan 3 at 18:17
Eevee Trainer
9,39331640
asked Jan 3 at 17:53
WorldGovWorldGov
345211
edited Jan 3 at 18:17
Eevee Trainer
9,39331640
edited Jan 3 at 18:17
Eevee Trainer
9,39331640
edited Jan 3 at 18:17
Eevee Trainer
9,39331640
9,39331640
asked Jan 3 at 17:53
WorldGovWorldGov
345211
asked Jan 3 at 17:53
WorldGovWorldGov
345211
asked Jan 3 at 17:53
WorldGovWorldGov
345211
345211
marked as duplicate by Martin R, Did, José Carlos Santos sequences-and-series
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Jan 3 at 17:59
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 Martin R, Did, José Carlos Santos sequences-and-series
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Jan 3 at 17:59
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.
$begingroup$
Looks like: $$sum_{r=0}^{infty}{frac{1}{4^r(2r+1)}}$$
$endgroup$
– Rhys Hughes
Jan 3 at 17:59
$begingroup$
The general term is $[(2n+1)4^n]^{-1}$ if that helps.
$endgroup$
– John
Jan 3 at 17:59
$begingroup$
The reason $log$ appears is simple: its derivative is $frac1x$, which is exactly what allows you to get all the odd numbers to appear in the denominators..
$endgroup$
– Arthur
Jan 3 at 18:00
3
$begingroup$
You should be careful about writing $+dotsinfty$, it looks like you're adding infinity to all the other fractions. This, however, is not what you mean.
$endgroup$
– Michael Burr
Jan 3 at 18:02
add a comment |
$begingroup$
Looks like: $$sum_{r=0}^{infty}{frac{1}{4^r(2r+1)}}$$
$endgroup$
– Rhys Hughes
Jan 3 at 17:59
$begingroup$
The general term is $[(2n+1)4^n]^{-1}$ if that helps.
$endgroup$
– John
Jan 3 at 17:59
$begingroup$
The reason $log$ appears is simple: its derivative is $frac1x$, which is exactly what allows you to get all the odd numbers to appear in the denominators..
$endgroup$
– Arthur
Jan 3 at 18:00
3
$begingroup$
You should be careful about writing $+dotsinfty$, it looks like you're adding infinity to all the other fractions. This, however, is not what you mean.
$endgroup$
– Michael Burr
Jan 3 at 18:02
$begingroup$
Looks like: $$sum_{r=0}^{infty}{frac{1}{4^r(2r+1)}}$$
$endgroup$
– Rhys Hughes
Jan 3 at 17:59
$begingroup$
Looks like: $$sum_{r=0}^{infty}{frac{1}{4^r(2r+1)}}$$
$endgroup$
– Rhys Hughes
Jan 3 at 17:59
$begingroup$
The general term is $[(2n+1)4^n]^{-1}$ if that helps.
$endgroup$
– John
Jan 3 at 17:59
$begingroup$
The general term is $[(2n+1)4^n]^{-1}$ if that helps.
$endgroup$
– John
Jan 3 at 17:59
$begingroup$
The reason $log$ appears is simple: its derivative is $frac1x$, which is exactly what allows you to get all the odd numbers to appear in the denominators..
$endgroup$
– Arthur
Jan 3 at 18:00
$begingroup$
The reason $log$ appears is simple: its derivative is $frac1x$, which is exactly what allows you to get all the odd numbers to appear in the denominators..
$endgroup$
– Arthur
Jan 3 at 18:00
3
3
$begingroup$
You should be careful about writing $+dotsinfty$, it looks like you're adding infinity to all the other fractions. This, however, is not what you mean.
$endgroup$
– Michael Burr
Jan 3 at 18:02
$begingroup$
You should be careful about writing $+dotsinfty$, it looks like you're adding infinity to all the other fractions. This, however, is not what you mean.
$endgroup$
– Michael Burr
Jan 3 at 18:02
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your series is:
$$f(x) = sum_{n=0}^inftydfrac{x^n}{2n+1}$$
with $x = dfrac{1}{4}.$ The standard way to go about this kind of problem is to observe that $f(x)$ only converges uniformly for $|x|leq r<1$ for any such $r$ and then manipulate the series into something we know already, usually involving taking derivatives.
If you want the challenge, then let me give you a hint: For $0<x<1$, consider $xf(x^2)$ and then take its derivative to get something you can easily calculate.
answered Jan 3 at 17:59
dezdichadodezdichado
6,4701929
$endgroup$
$begingroup$
@J.G. you are right, that was a mistake.
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– dezdichado
Jan 3 at 18:01
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your series is:
$$f(x) = sum_{n=0}^inftydfrac{x^n}{2n+1}$$
with $x = dfrac{1}{4}.$ The standard way to go about this kind of problem is to observe that $f(x)$ only converges uniformly for $|x|leq r<1$ for any such $r$ and then manipulate the series into something we know already, usually involving taking derivatives.
If you want the challenge, then let me give you a hint: For $0<x<1$, consider $xf(x^2)$ and then take its derivative to get something you can easily calculate.
answered Jan 3 at 17:59
dezdichadodezdichado
6,4701929
$endgroup$
$begingroup$
@J.G. you are right, that was a mistake.
$endgroup$
– dezdichado
Jan 3 at 18:01
add a comment |
$begingroup$
Your series is:
$$f(x) = sum_{n=0}^inftydfrac{x^n}{2n+1}$$
with $x = dfrac{1}{4}.$ The standard way to go about this kind of problem is to observe that $f(x)$ only converges uniformly for $|x|leq r<1$ for any such $r$ and then manipulate the series into something we know already, usually involving taking derivatives.
If you want the challenge, then let me give you a hint: For $0<x<1$, consider $xf(x^2)$ and then take its derivative to get something you can easily calculate.
answered Jan 3 at 17:59
dezdichadodezdichado
6,4701929
$endgroup$
$begingroup$
@J.G. you are right, that was a mistake.
$endgroup$
– dezdichado
Jan 3 at 18:01
add a comment |
$begingroup$
Your series is:
$$f(x) = sum_{n=0}^inftydfrac{x^n}{2n+1}$$
with $x = dfrac{1}{4}.$ The standard way to go about this kind of problem is to observe that $f(x)$ only converges uniformly for $|x|leq r<1$ for any such $r$ and then manipulate the series into something we know already, usually involving taking derivatives.
If you want the challenge, then let me give you a hint: For $0<x<1$, consider $xf(x^2)$ and then take its derivative to get something you can easily calculate.
answered Jan 3 at 17:59
dezdichadodezdichado
6,4701929
$endgroup$
Your series is:
$$f(x) = sum_{n=0}^inftydfrac{x^n}{2n+1}$$
with $x = dfrac{1}{4}.$ The standard way to go about this kind of problem is to observe that $f(x)$ only converges uniformly for $|x|leq r<1$ for any such $r$ and then manipulate the series into something we know already, usually involving taking derivatives.
If you want the challenge, then let me give you a hint: For $0<x<1$, consider $xf(x^2)$ and then take its derivative to get something you can easily calculate.
answered Jan 3 at 17:59
dezdichadodezdichado
6,4701929
edited Jan 3 at 18:01
answered Jan 3 at 17:59
dezdichadodezdichado
6,4701929
answered Jan 3 at 17:59
dezdichadodezdichado
6,4701929
answered Jan 3 at 17:59
dezdichadodezdichado
6,4701929
6,4701929
$begingroup$
@J.G. you are right, that was a mistake.
$endgroup$
– dezdichado
Jan 3 at 18:01
add a comment |
$begingroup$
@J.G. you are right, that was a mistake.
$endgroup$
– dezdichado
Jan 3 at 18:01
$begingroup$
@J.G. you are right, that was a mistake.
$endgroup$
– dezdichado
Jan 3 at 18:01
$begingroup$
@J.G. you are right, that was a mistake.
$endgroup$
– dezdichado
Jan 3 at 18:01
add a comment |
$begingroup$
Looks like: $$sum_{r=0}^{infty}{frac{1}{4^r(2r+1)}}$$
$endgroup$
– Rhys Hughes
Jan 3 at 17:59
$begingroup$
The general term is $[(2n+1)4^n]^{-1}$ if that helps.
$endgroup$
– John
Jan 3 at 17:59
$begingroup$
The reason $log$ appears is simple: its derivative is $frac1x$, which is exactly what allows you to get all the odd numbers to appear in the denominators..
$endgroup$
– Arthur
Jan 3 at 18:00
3
$begingroup$
You should be careful about writing $+dotsinfty$, it looks like you're adding infinity to all the other fractions. This, however, is not what you mean.
$endgroup$
– Michael Burr
Jan 3 at 18:02