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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  • 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







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marked as duplicate by Martin R, Did, José Carlos Santos sequences-and-series
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 2 days ago


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.















  • Looks like: $$sum_{r=0}^{infty}{frac{1}{4^r(2r+1)}}$$
    – Rhys Hughes
    2 days ago












  • The general term is $[(2n+1)4^n]^{-1}$ if that helps.
    – John
    2 days ago










  • 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..
    – Arthur
    2 days ago








  • 3




    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.
    – Michael Burr
    2 days ago


















0















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







share|cite|improve this question















marked as duplicate by Martin R, Did, José Carlos Santos sequences-and-series
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 2 days ago


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.















  • Looks like: $$sum_{r=0}^{infty}{frac{1}{4^r(2r+1)}}$$
    – Rhys Hughes
    2 days ago












  • The general term is $[(2n+1)4^n]^{-1}$ if that helps.
    – John
    2 days ago










  • 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..
    – Arthur
    2 days ago








  • 3




    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.
    – Michael Burr
    2 days ago
















0












0








0








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







share|cite|improve this question
















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Eevee Trainer

4,9321634




4,9321634










asked 2 days ago









WorldGov

2629




2629




marked as duplicate by Martin R, Did, José Carlos Santos sequences-and-series
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 2 days ago


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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 2 days ago


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.














  • Looks like: $$sum_{r=0}^{infty}{frac{1}{4^r(2r+1)}}$$
    – Rhys Hughes
    2 days ago












  • The general term is $[(2n+1)4^n]^{-1}$ if that helps.
    – John
    2 days ago










  • 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..
    – Arthur
    2 days ago








  • 3




    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.
    – Michael Burr
    2 days ago




















  • Looks like: $$sum_{r=0}^{infty}{frac{1}{4^r(2r+1)}}$$
    – Rhys Hughes
    2 days ago












  • The general term is $[(2n+1)4^n]^{-1}$ if that helps.
    – John
    2 days ago










  • 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..
    – Arthur
    2 days ago








  • 3




    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.
    – Michael Burr
    2 days ago


















Looks like: $$sum_{r=0}^{infty}{frac{1}{4^r(2r+1)}}$$
– Rhys Hughes
2 days ago






Looks like: $$sum_{r=0}^{infty}{frac{1}{4^r(2r+1)}}$$
– Rhys Hughes
2 days ago














The general term is $[(2n+1)4^n]^{-1}$ if that helps.
– John
2 days ago




The general term is $[(2n+1)4^n]^{-1}$ if that helps.
– John
2 days ago












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..
– Arthur
2 days ago






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..
– Arthur
2 days ago






3




3




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.
– Michael Burr
2 days ago






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.
– Michael Burr
2 days ago












1 Answer
1






active

oldest

votes


















4














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.






share|cite|improve this answer























  • @J.G. you are right, that was a mistake.
    – dezdichado
    2 days ago


















1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4














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.






share|cite|improve this answer























  • @J.G. you are right, that was a mistake.
    – dezdichado
    2 days ago
















4














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.






share|cite|improve this answer























  • @J.G. you are right, that was a mistake.
    – dezdichado
    2 days ago














4












4








4






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.






share|cite|improve this answer














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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 days ago

























answered 2 days ago









dezdichado

6,2551929




6,2551929












  • @J.G. you are right, that was a mistake.
    – dezdichado
    2 days ago


















  • @J.G. you are right, that was a mistake.
    – dezdichado
    2 days ago
















@J.G. you are right, that was a mistake.
– dezdichado
2 days ago




@J.G. you are right, that was a mistake.
– dezdichado
2 days ago



-

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