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Closed form representation of alternating series


Alternating seriesConvergence of alternating seriesClosed-forms of infinite series with factorial in the denominatorHow to evaluate a partially alternating series.Closed form representation of an exponential seriesAlternating series test for complex seriesAlternating series that is bounded by 1 but fails an alternating series testrearrangement of the alternating harmonic seriesWhat is the notation for alternating series with “$cdots$”?Upper bound on alternating harmonic series













1












$begingroup$


I consider the following series
begin{align}
sum_{n=0}^infty{frac{(-1)^{n+2}}{n+2}} stackrel{?}{=} 1-ln(2)
end{align}
Wolfram tells me that it is equal to $1-ln(2)$.
I know the following
begin{align}
&sum_{k=1}^infty{frac{(-1)^{k+1}}{k}} = ln(2)\
&sum_{n=0}^infty{frac{(-1)^{n+2}}{n+2}} = sum_{k=1}^infty{frac{(-1)^{k+1}}{k+1}}
end{align}
I thus need to show that the following is true, but I would not know how to do it.
begin{align}
sum_{k=1}^infty{frac{(-1)^{k+1}}{k+1}} + sum_{k=1}^infty{frac{(-1)^{k+1}}{k}} = 1
end{align}






sequences-and-series







share|cite|improve this question









$endgroup$












  • $begingroup$
    if you put both series into one and start computing the first terms, you will notice that something nice is happening
    $endgroup$
    – J.F
    Mar 15 at 10:07
















1












$begingroup$


I consider the following series
begin{align}
sum_{n=0}^infty{frac{(-1)^{n+2}}{n+2}} stackrel{?}{=} 1-ln(2)
end{align}
Wolfram tells me that it is equal to $1-ln(2)$.
I know the following
begin{align}
&sum_{k=1}^infty{frac{(-1)^{k+1}}{k}} = ln(2)\
&sum_{n=0}^infty{frac{(-1)^{n+2}}{n+2}} = sum_{k=1}^infty{frac{(-1)^{k+1}}{k+1}}
end{align}
I thus need to show that the following is true, but I would not know how to do it.
begin{align}
sum_{k=1}^infty{frac{(-1)^{k+1}}{k+1}} + sum_{k=1}^infty{frac{(-1)^{k+1}}{k}} = 1
end{align}






sequences-and-series







share|cite|improve this question









$endgroup$












  • $begingroup$
    if you put both series into one and start computing the first terms, you will notice that something nice is happening
    $endgroup$
    – J.F
    Mar 15 at 10:07














1












1








1





$begingroup$


I consider the following series
begin{align}
sum_{n=0}^infty{frac{(-1)^{n+2}}{n+2}} stackrel{?}{=} 1-ln(2)
end{align}
Wolfram tells me that it is equal to $1-ln(2)$.
I know the following
begin{align}
&sum_{k=1}^infty{frac{(-1)^{k+1}}{k}} = ln(2)\
&sum_{n=0}^infty{frac{(-1)^{n+2}}{n+2}} = sum_{k=1}^infty{frac{(-1)^{k+1}}{k+1}}
end{align}
I thus need to show that the following is true, but I would not know how to do it.
begin{align}
sum_{k=1}^infty{frac{(-1)^{k+1}}{k+1}} + sum_{k=1}^infty{frac{(-1)^{k+1}}{k}} = 1
end{align}






sequences-and-series







share|cite|improve this question









$endgroup$




I consider the following series
begin{align}
sum_{n=0}^infty{frac{(-1)^{n+2}}{n+2}} stackrel{?}{=} 1-ln(2)
end{align}
Wolfram tells me that it is equal to $1-ln(2)$.
I know the following
begin{align}
&sum_{k=1}^infty{frac{(-1)^{k+1}}{k}} = ln(2)\
&sum_{n=0}^infty{frac{(-1)^{n+2}}{n+2}} = sum_{k=1}^infty{frac{(-1)^{k+1}}{k+1}}
end{align}
I thus need to show that the following is true, but I would not know how to do it.
begin{align}
sum_{k=1}^infty{frac{(-1)^{k+1}}{k+1}} + sum_{k=1}^infty{frac{(-1)^{k+1}}{k}} = 1
end{align}






sequences-and-series




sequences-and-series






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 15 at 10:05









cluelessclueless

307111




307111












  • $begingroup$
    if you put both series into one and start computing the first terms, you will notice that something nice is happening
    $endgroup$
    – J.F
    Mar 15 at 10:07


















  • $begingroup$
    if you put both series into one and start computing the first terms, you will notice that something nice is happening
    $endgroup$
    – J.F
    Mar 15 at 10:07
















$begingroup$
if you put both series into one and start computing the first terms, you will notice that something nice is happening
$endgroup$
– J.F
Mar 15 at 10:07




$begingroup$
if you put both series into one and start computing the first terms, you will notice that something nice is happening
$endgroup$
– J.F
Mar 15 at 10:07










3 Answers
3






active

oldest

votes


















1












$begingroup$

$sum_{n=0}^{infty}dfrac{(-1)^{n+2}}{n+2}=$



$sum_{k=2}^{infty}dfrac{(-1)^k}{k} -1+1=$



$sum_{k=1}^{infty}dfrac{(-1)^k}{k} +1=$



$1+ (-1)sum_{k=1}^{infty}dfrac{(-1)^{k+1}}{k}=$



$1- log 2.$






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    begin{align}
    sum_{k=1}^infty{frac{(-1)^{k+1}}{k+1}} + sum_{k=1}^infty{frac{(-1)^{k+1}}{k}}
    &=
    sum_{k=1}^infty{frac{(-1)^{k+1}}{k+1}} + sum_{k=0}^infty{frac{(-1)^{k}}{k+1}}
    \ &=
    sum_{k=1}^infty{frac{(-1)^{k+1}}{k+1}} + sum_{k=1}^infty{frac{(-1)^{k}}{k+1}}+1
    \ &=1
    end{align}






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      $sum_{k=1}^infty{frac{(-1)^{k+1}}{k+1}}=frac 1 2-frac 1 3+frac 1 4+cdots$ and $sum_{k=1}^infty{frac{(-1)^{k+1}}{k}}=1-frac 1 2+frac 1 3-frac 1 4+cdots$. What happens when you add these two?






      share|cite|improve this answer









      $endgroup$













        Your Answer





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        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1












        $begingroup$

        $sum_{n=0}^{infty}dfrac{(-1)^{n+2}}{n+2}=$



        $sum_{k=2}^{infty}dfrac{(-1)^k}{k} -1+1=$



        $sum_{k=1}^{infty}dfrac{(-1)^k}{k} +1=$



        $1+ (-1)sum_{k=1}^{infty}dfrac{(-1)^{k+1}}{k}=$



        $1- log 2.$






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          $sum_{n=0}^{infty}dfrac{(-1)^{n+2}}{n+2}=$



          $sum_{k=2}^{infty}dfrac{(-1)^k}{k} -1+1=$



          $sum_{k=1}^{infty}dfrac{(-1)^k}{k} +1=$



          $1+ (-1)sum_{k=1}^{infty}dfrac{(-1)^{k+1}}{k}=$



          $1- log 2.$






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            $sum_{n=0}^{infty}dfrac{(-1)^{n+2}}{n+2}=$



            $sum_{k=2}^{infty}dfrac{(-1)^k}{k} -1+1=$



            $sum_{k=1}^{infty}dfrac{(-1)^k}{k} +1=$



            $1+ (-1)sum_{k=1}^{infty}dfrac{(-1)^{k+1}}{k}=$



            $1- log 2.$






            share|cite|improve this answer









            $endgroup$



            $sum_{n=0}^{infty}dfrac{(-1)^{n+2}}{n+2}=$



            $sum_{k=2}^{infty}dfrac{(-1)^k}{k} -1+1=$



            $sum_{k=1}^{infty}dfrac{(-1)^k}{k} +1=$



            $1+ (-1)sum_{k=1}^{infty}dfrac{(-1)^{k+1}}{k}=$



            $1- log 2.$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Mar 15 at 10:34









            Peter SzilasPeter Szilas

            11.6k2822




            11.6k2822























                2












                $begingroup$

                begin{align}
                sum_{k=1}^infty{frac{(-1)^{k+1}}{k+1}} + sum_{k=1}^infty{frac{(-1)^{k+1}}{k}}
                &=
                sum_{k=1}^infty{frac{(-1)^{k+1}}{k+1}} + sum_{k=0}^infty{frac{(-1)^{k}}{k+1}}
                \ &=
                sum_{k=1}^infty{frac{(-1)^{k+1}}{k+1}} + sum_{k=1}^infty{frac{(-1)^{k}}{k+1}}+1
                \ &=1
                end{align}






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  begin{align}
                  sum_{k=1}^infty{frac{(-1)^{k+1}}{k+1}} + sum_{k=1}^infty{frac{(-1)^{k+1}}{k}}
                  &=
                  sum_{k=1}^infty{frac{(-1)^{k+1}}{k+1}} + sum_{k=0}^infty{frac{(-1)^{k}}{k+1}}
                  \ &=
                  sum_{k=1}^infty{frac{(-1)^{k+1}}{k+1}} + sum_{k=1}^infty{frac{(-1)^{k}}{k+1}}+1
                  \ &=1
                  end{align}






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    begin{align}
                    sum_{k=1}^infty{frac{(-1)^{k+1}}{k+1}} + sum_{k=1}^infty{frac{(-1)^{k+1}}{k}}
                    &=
                    sum_{k=1}^infty{frac{(-1)^{k+1}}{k+1}} + sum_{k=0}^infty{frac{(-1)^{k}}{k+1}}
                    \ &=
                    sum_{k=1}^infty{frac{(-1)^{k+1}}{k+1}} + sum_{k=1}^infty{frac{(-1)^{k}}{k+1}}+1
                    \ &=1
                    end{align}






                    share|cite|improve this answer









                    $endgroup$



                    begin{align}
                    sum_{k=1}^infty{frac{(-1)^{k+1}}{k+1}} + sum_{k=1}^infty{frac{(-1)^{k+1}}{k}}
                    &=
                    sum_{k=1}^infty{frac{(-1)^{k+1}}{k+1}} + sum_{k=0}^infty{frac{(-1)^{k}}{k+1}}
                    \ &=
                    sum_{k=1}^infty{frac{(-1)^{k+1}}{k+1}} + sum_{k=1}^infty{frac{(-1)^{k}}{k+1}}+1
                    \ &=1
                    end{align}







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Mar 15 at 10:08









                    ZeroZero

                    50311




                    50311























                        1












                        $begingroup$

                        $sum_{k=1}^infty{frac{(-1)^{k+1}}{k+1}}=frac 1 2-frac 1 3+frac 1 4+cdots$ and $sum_{k=1}^infty{frac{(-1)^{k+1}}{k}}=1-frac 1 2+frac 1 3-frac 1 4+cdots$. What happens when you add these two?






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          $sum_{k=1}^infty{frac{(-1)^{k+1}}{k+1}}=frac 1 2-frac 1 3+frac 1 4+cdots$ and $sum_{k=1}^infty{frac{(-1)^{k+1}}{k}}=1-frac 1 2+frac 1 3-frac 1 4+cdots$. What happens when you add these two?






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            $sum_{k=1}^infty{frac{(-1)^{k+1}}{k+1}}=frac 1 2-frac 1 3+frac 1 4+cdots$ and $sum_{k=1}^infty{frac{(-1)^{k+1}}{k}}=1-frac 1 2+frac 1 3-frac 1 4+cdots$. What happens when you add these two?






                            share|cite|improve this answer









                            $endgroup$



                            $sum_{k=1}^infty{frac{(-1)^{k+1}}{k+1}}=frac 1 2-frac 1 3+frac 1 4+cdots$ and $sum_{k=1}^infty{frac{(-1)^{k+1}}{k}}=1-frac 1 2+frac 1 3-frac 1 4+cdots$. What happens when you add these two?







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Mar 15 at 10:13









                            Kavi Rama MurthyKavi Rama Murthy

                            70.2k53170




                            70.2k53170






























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