Evaluating the Cauchy product of $sum_{n=0}^{infty}frac{(-1)^{n+1}}{n+1}$ and $sum_{n=0}^{infty}frac{1}{3^n}...












4












$begingroup$


Using the Mertens' theorem for Cauchy products we know that the Cauchy product of series
$$
sum_{n=0}^{infty}frac{(-1)^{n+1}}{n+1}qquadtext{and}qquadsum_{n=0}^{infty}frac{1}{3^n}
$$ does converge. But how can we try to find the value of this sum?



What I only know is that we can write that this sum equals to
$$
S=sum_{n=0}^{infty}c_nqquadtext{where}qquad c_n=sum_{k=0}^{n}frac{1}{3^{n-k}}frac{(-1)^{k+1}}{k+1}
$$
but I'm not able to find any way to calculate the sum. Any hints?






sequences-and-series summation







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    If I understand correctly, by MERTENS' Theorem, $S$ is equal to the product of the of the initial sums, right? The first is $-ln(2)$ and the second is $frac{3}{2}$ by the geometric series.
    $endgroup$
    – Viktor Glombik
    Dec 23 '18 at 12:38








  • 1




    $begingroup$
    Yes but how to find it using only the definition of Cauchy product of series?
    $endgroup$
    – avan1235
    Dec 23 '18 at 12:44










  • $begingroup$
    If you were presented with this sum as a $S$ and asked to evaluate it. I would think that decomposing it (via cauchy product) would be the easiest way of coming to a conclusion.
    $endgroup$
    – Mason
    Dec 23 '18 at 16:11
















4












$begingroup$


Using the Mertens' theorem for Cauchy products we know that the Cauchy product of series
$$
sum_{n=0}^{infty}frac{(-1)^{n+1}}{n+1}qquadtext{and}qquadsum_{n=0}^{infty}frac{1}{3^n}
$$ does converge. But how can we try to find the value of this sum?



What I only know is that we can write that this sum equals to
$$
S=sum_{n=0}^{infty}c_nqquadtext{where}qquad c_n=sum_{k=0}^{n}frac{1}{3^{n-k}}frac{(-1)^{k+1}}{k+1}
$$
but I'm not able to find any way to calculate the sum. Any hints?






sequences-and-series summation







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    If I understand correctly, by MERTENS' Theorem, $S$ is equal to the product of the of the initial sums, right? The first is $-ln(2)$ and the second is $frac{3}{2}$ by the geometric series.
    $endgroup$
    – Viktor Glombik
    Dec 23 '18 at 12:38








  • 1




    $begingroup$
    Yes but how to find it using only the definition of Cauchy product of series?
    $endgroup$
    – avan1235
    Dec 23 '18 at 12:44










  • $begingroup$
    If you were presented with this sum as a $S$ and asked to evaluate it. I would think that decomposing it (via cauchy product) would be the easiest way of coming to a conclusion.
    $endgroup$
    – Mason
    Dec 23 '18 at 16:11














4












4








4


0



$begingroup$


Using the Mertens' theorem for Cauchy products we know that the Cauchy product of series
$$
sum_{n=0}^{infty}frac{(-1)^{n+1}}{n+1}qquadtext{and}qquadsum_{n=0}^{infty}frac{1}{3^n}
$$ does converge. But how can we try to find the value of this sum?



What I only know is that we can write that this sum equals to
$$
S=sum_{n=0}^{infty}c_nqquadtext{where}qquad c_n=sum_{k=0}^{n}frac{1}{3^{n-k}}frac{(-1)^{k+1}}{k+1}
$$
but I'm not able to find any way to calculate the sum. Any hints?






sequences-and-series summation







share|cite|improve this question











$endgroup$




Using the Mertens' theorem for Cauchy products we know that the Cauchy product of series
$$
sum_{n=0}^{infty}frac{(-1)^{n+1}}{n+1}qquadtext{and}qquadsum_{n=0}^{infty}frac{1}{3^n}
$$ does converge. But how can we try to find the value of this sum?



What I only know is that we can write that this sum equals to
$$
S=sum_{n=0}^{infty}c_nqquadtext{where}qquad c_n=sum_{k=0}^{n}frac{1}{3^{n-k}}frac{(-1)^{k+1}}{k+1}
$$
but I'm not able to find any way to calculate the sum. Any hints?






sequences-and-series summation




sequences-and-series summation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 23 '18 at 12:30









Blue

49.3k870157




49.3k870157










asked Dec 23 '18 at 12:23









avan1235avan1235

3578




3578








  • 5




    $begingroup$
    If I understand correctly, by MERTENS' Theorem, $S$ is equal to the product of the of the initial sums, right? The first is $-ln(2)$ and the second is $frac{3}{2}$ by the geometric series.
    $endgroup$
    – Viktor Glombik
    Dec 23 '18 at 12:38








  • 1




    $begingroup$
    Yes but how to find it using only the definition of Cauchy product of series?
    $endgroup$
    – avan1235
    Dec 23 '18 at 12:44










  • $begingroup$
    If you were presented with this sum as a $S$ and asked to evaluate it. I would think that decomposing it (via cauchy product) would be the easiest way of coming to a conclusion.
    $endgroup$
    – Mason
    Dec 23 '18 at 16:11














  • 5




    $begingroup$
    If I understand correctly, by MERTENS' Theorem, $S$ is equal to the product of the of the initial sums, right? The first is $-ln(2)$ and the second is $frac{3}{2}$ by the geometric series.
    $endgroup$
    – Viktor Glombik
    Dec 23 '18 at 12:38








  • 1




    $begingroup$
    Yes but how to find it using only the definition of Cauchy product of series?
    $endgroup$
    – avan1235
    Dec 23 '18 at 12:44










  • $begingroup$
    If you were presented with this sum as a $S$ and asked to evaluate it. I would think that decomposing it (via cauchy product) would be the easiest way of coming to a conclusion.
    $endgroup$
    – Mason
    Dec 23 '18 at 16:11








5




5




$begingroup$
If I understand correctly, by MERTENS' Theorem, $S$ is equal to the product of the of the initial sums, right? The first is $-ln(2)$ and the second is $frac{3}{2}$ by the geometric series.
$endgroup$
– Viktor Glombik
Dec 23 '18 at 12:38






$begingroup$
If I understand correctly, by MERTENS' Theorem, $S$ is equal to the product of the of the initial sums, right? The first is $-ln(2)$ and the second is $frac{3}{2}$ by the geometric series.
$endgroup$
– Viktor Glombik
Dec 23 '18 at 12:38






1




1




$begingroup$
Yes but how to find it using only the definition of Cauchy product of series?
$endgroup$
– avan1235
Dec 23 '18 at 12:44




$begingroup$
Yes but how to find it using only the definition of Cauchy product of series?
$endgroup$
– avan1235
Dec 23 '18 at 12:44












$begingroup$
If you were presented with this sum as a $S$ and asked to evaluate it. I would think that decomposing it (via cauchy product) would be the easiest way of coming to a conclusion.
$endgroup$
– Mason
Dec 23 '18 at 16:11




$begingroup$
If you were presented with this sum as a $S$ and asked to evaluate it. I would think that decomposing it (via cauchy product) would be the easiest way of coming to a conclusion.
$endgroup$
– Mason
Dec 23 '18 at 16:11










1 Answer
1






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oldest

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3












$begingroup$

Here is my "lengthy" solution (elaborating @ViktorGlombik comment). Let's start with your sum:



$$S=sum_{n=0}^inftyfrac{1}{3^n}sum_{k=0}^nfrac{1}{3^{-k}}frac{(-1)^{k+1}}{k+1}$$



where



$$sum_{k=0}^nfrac{1}{3^{-k}}frac{(-1)^{k+1}}{k+1}=(-3)^{n+1}Phi(-3,1,n+2)-frac{2}{3}log(2)$$



where $Phi(a,b,c)$ is so-called Lerch transcendent.



Also there is one useful identity: $Phi(-3, 1, n+2) = frac{1}{Gamma(1)}int_0^inftyfrac{e^{-(n+2)t}}{1+3e^{-t}}dt$.



By substitution $u=1+3e^{-t}$ (and $Gamma(1) = 1$), this integral leads to



$$Phi(-3, 1, n+2) = frac{1}{3^{n+2}}int_1^4 frac{(u-1)^{n+1}}{u} du.$$



Plugging it back (observe $uin(1,4)$ for the convergence of the geometric series), we obtain:
$$S=sum_{n=0}^inftyfrac{1}{3^n}left[(-3)^{n+1}frac{1}{3^{n+2}}int_1^4frac{(u-1)^{n+1}}{u}du-frac{2}{3}log(2)right]=$$
$$=sum_{n=0}^inftyfrac{1}{3^{n+1}}int_1^4frac{(1-u)^{n+1}}{u}du-log(2)=$$
$$=int_1^4frac{1}{u}sum_{n=0}^{infty}left(frac{1-u}{3}right)^{n+1}du - log(2)=$$
$$=int_1^4frac{1}{u}frac{1-u}{u+2}du-log(2)=-frac{3}{2}log(2)+log(2)-log(2)=-frac{3}{2}log(2).$$



So yes, indeed the sum converges to $S=-frac{3}{2}log(2)$.






share|cite|improve this answer











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    1 Answer
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    active

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    $begingroup$

    Here is my "lengthy" solution (elaborating @ViktorGlombik comment). Let's start with your sum:



    $$S=sum_{n=0}^inftyfrac{1}{3^n}sum_{k=0}^nfrac{1}{3^{-k}}frac{(-1)^{k+1}}{k+1}$$



    where



    $$sum_{k=0}^nfrac{1}{3^{-k}}frac{(-1)^{k+1}}{k+1}=(-3)^{n+1}Phi(-3,1,n+2)-frac{2}{3}log(2)$$



    where $Phi(a,b,c)$ is so-called Lerch transcendent.



    Also there is one useful identity: $Phi(-3, 1, n+2) = frac{1}{Gamma(1)}int_0^inftyfrac{e^{-(n+2)t}}{1+3e^{-t}}dt$.



    By substitution $u=1+3e^{-t}$ (and $Gamma(1) = 1$), this integral leads to



    $$Phi(-3, 1, n+2) = frac{1}{3^{n+2}}int_1^4 frac{(u-1)^{n+1}}{u} du.$$



    Plugging it back (observe $uin(1,4)$ for the convergence of the geometric series), we obtain:
    $$S=sum_{n=0}^inftyfrac{1}{3^n}left[(-3)^{n+1}frac{1}{3^{n+2}}int_1^4frac{(u-1)^{n+1}}{u}du-frac{2}{3}log(2)right]=$$
    $$=sum_{n=0}^inftyfrac{1}{3^{n+1}}int_1^4frac{(1-u)^{n+1}}{u}du-log(2)=$$
    $$=int_1^4frac{1}{u}sum_{n=0}^{infty}left(frac{1-u}{3}right)^{n+1}du - log(2)=$$
    $$=int_1^4frac{1}{u}frac{1-u}{u+2}du-log(2)=-frac{3}{2}log(2)+log(2)-log(2)=-frac{3}{2}log(2).$$



    So yes, indeed the sum converges to $S=-frac{3}{2}log(2)$.






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      Here is my "lengthy" solution (elaborating @ViktorGlombik comment). Let's start with your sum:



      $$S=sum_{n=0}^inftyfrac{1}{3^n}sum_{k=0}^nfrac{1}{3^{-k}}frac{(-1)^{k+1}}{k+1}$$



      where



      $$sum_{k=0}^nfrac{1}{3^{-k}}frac{(-1)^{k+1}}{k+1}=(-3)^{n+1}Phi(-3,1,n+2)-frac{2}{3}log(2)$$



      where $Phi(a,b,c)$ is so-called Lerch transcendent.



      Also there is one useful identity: $Phi(-3, 1, n+2) = frac{1}{Gamma(1)}int_0^inftyfrac{e^{-(n+2)t}}{1+3e^{-t}}dt$.



      By substitution $u=1+3e^{-t}$ (and $Gamma(1) = 1$), this integral leads to



      $$Phi(-3, 1, n+2) = frac{1}{3^{n+2}}int_1^4 frac{(u-1)^{n+1}}{u} du.$$



      Plugging it back (observe $uin(1,4)$ for the convergence of the geometric series), we obtain:
      $$S=sum_{n=0}^inftyfrac{1}{3^n}left[(-3)^{n+1}frac{1}{3^{n+2}}int_1^4frac{(u-1)^{n+1}}{u}du-frac{2}{3}log(2)right]=$$
      $$=sum_{n=0}^inftyfrac{1}{3^{n+1}}int_1^4frac{(1-u)^{n+1}}{u}du-log(2)=$$
      $$=int_1^4frac{1}{u}sum_{n=0}^{infty}left(frac{1-u}{3}right)^{n+1}du - log(2)=$$
      $$=int_1^4frac{1}{u}frac{1-u}{u+2}du-log(2)=-frac{3}{2}log(2)+log(2)-log(2)=-frac{3}{2}log(2).$$



      So yes, indeed the sum converges to $S=-frac{3}{2}log(2)$.






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        Here is my "lengthy" solution (elaborating @ViktorGlombik comment). Let's start with your sum:



        $$S=sum_{n=0}^inftyfrac{1}{3^n}sum_{k=0}^nfrac{1}{3^{-k}}frac{(-1)^{k+1}}{k+1}$$



        where



        $$sum_{k=0}^nfrac{1}{3^{-k}}frac{(-1)^{k+1}}{k+1}=(-3)^{n+1}Phi(-3,1,n+2)-frac{2}{3}log(2)$$



        where $Phi(a,b,c)$ is so-called Lerch transcendent.



        Also there is one useful identity: $Phi(-3, 1, n+2) = frac{1}{Gamma(1)}int_0^inftyfrac{e^{-(n+2)t}}{1+3e^{-t}}dt$.



        By substitution $u=1+3e^{-t}$ (and $Gamma(1) = 1$), this integral leads to



        $$Phi(-3, 1, n+2) = frac{1}{3^{n+2}}int_1^4 frac{(u-1)^{n+1}}{u} du.$$



        Plugging it back (observe $uin(1,4)$ for the convergence of the geometric series), we obtain:
        $$S=sum_{n=0}^inftyfrac{1}{3^n}left[(-3)^{n+1}frac{1}{3^{n+2}}int_1^4frac{(u-1)^{n+1}}{u}du-frac{2}{3}log(2)right]=$$
        $$=sum_{n=0}^inftyfrac{1}{3^{n+1}}int_1^4frac{(1-u)^{n+1}}{u}du-log(2)=$$
        $$=int_1^4frac{1}{u}sum_{n=0}^{infty}left(frac{1-u}{3}right)^{n+1}du - log(2)=$$
        $$=int_1^4frac{1}{u}frac{1-u}{u+2}du-log(2)=-frac{3}{2}log(2)+log(2)-log(2)=-frac{3}{2}log(2).$$



        So yes, indeed the sum converges to $S=-frac{3}{2}log(2)$.






        share|cite|improve this answer











        $endgroup$



        Here is my "lengthy" solution (elaborating @ViktorGlombik comment). Let's start with your sum:



        $$S=sum_{n=0}^inftyfrac{1}{3^n}sum_{k=0}^nfrac{1}{3^{-k}}frac{(-1)^{k+1}}{k+1}$$



        where



        $$sum_{k=0}^nfrac{1}{3^{-k}}frac{(-1)^{k+1}}{k+1}=(-3)^{n+1}Phi(-3,1,n+2)-frac{2}{3}log(2)$$



        where $Phi(a,b,c)$ is so-called Lerch transcendent.



        Also there is one useful identity: $Phi(-3, 1, n+2) = frac{1}{Gamma(1)}int_0^inftyfrac{e^{-(n+2)t}}{1+3e^{-t}}dt$.



        By substitution $u=1+3e^{-t}$ (and $Gamma(1) = 1$), this integral leads to



        $$Phi(-3, 1, n+2) = frac{1}{3^{n+2}}int_1^4 frac{(u-1)^{n+1}}{u} du.$$



        Plugging it back (observe $uin(1,4)$ for the convergence of the geometric series), we obtain:
        $$S=sum_{n=0}^inftyfrac{1}{3^n}left[(-3)^{n+1}frac{1}{3^{n+2}}int_1^4frac{(u-1)^{n+1}}{u}du-frac{2}{3}log(2)right]=$$
        $$=sum_{n=0}^inftyfrac{1}{3^{n+1}}int_1^4frac{(1-u)^{n+1}}{u}du-log(2)=$$
        $$=int_1^4frac{1}{u}sum_{n=0}^{infty}left(frac{1-u}{3}right)^{n+1}du - log(2)=$$
        $$=int_1^4frac{1}{u}frac{1-u}{u+2}du-log(2)=-frac{3}{2}log(2)+log(2)-log(2)=-frac{3}{2}log(2).$$



        So yes, indeed the sum converges to $S=-frac{3}{2}log(2)$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 25 '18 at 9:18

























        answered Dec 23 '18 at 22:15









        pisoirpisoir

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