Calculating the limit $lim_{ntoinfty}[sum_{k=0}^{n-1}(-1)^kfrac{1}{n-k}]$












1












$begingroup$


So, I have got the series: $$I_n = int_0^1frac{x^n}{1+x}dx$$ and my task is to find the limit
$$lim_{ntoinfty}(I_n)$$
I have added 1 and subtracted 1 from the numerator and factorised $x^n - 1$ as $(x+1)(x^{n-1}-x^{n-2}+x^{n-3} - ...+(-1)^n)$. Then i did the reduction, and I'm left with calculating the limit:
$$lim_{ntoinfty}[frac{1}{n}-frac{1}{n-1}+frac{1}{n-2} -...+(-1)^{n-1}frac{1}{2} + (-1)^n] = lim_{ntoinfty}[sum_{k=0}^{n-1}(-1)^kfrac{1}{n-k}]$$
I know I can calculate the fraction as a Riemann Sum using the integral of the function $frac{1}{1-x}$, however I don't know how to deal with the changing signs. How should i approach this? Is there some sort of rule or common practice to deal with alternating signs?






sequences-and-series limits







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    1












    $begingroup$


    So, I have got the series: $$I_n = int_0^1frac{x^n}{1+x}dx$$ and my task is to find the limit
    $$lim_{ntoinfty}(I_n)$$
    I have added 1 and subtracted 1 from the numerator and factorised $x^n - 1$ as $(x+1)(x^{n-1}-x^{n-2}+x^{n-3} - ...+(-1)^n)$. Then i did the reduction, and I'm left with calculating the limit:
    $$lim_{ntoinfty}[frac{1}{n}-frac{1}{n-1}+frac{1}{n-2} -...+(-1)^{n-1}frac{1}{2} + (-1)^n] = lim_{ntoinfty}[sum_{k=0}^{n-1}(-1)^kfrac{1}{n-k}]$$
    I know I can calculate the fraction as a Riemann Sum using the integral of the function $frac{1}{1-x}$, however I don't know how to deal with the changing signs. How should i approach this? Is there some sort of rule or common practice to deal with alternating signs?






    sequences-and-series limits







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      So, I have got the series: $$I_n = int_0^1frac{x^n}{1+x}dx$$ and my task is to find the limit
      $$lim_{ntoinfty}(I_n)$$
      I have added 1 and subtracted 1 from the numerator and factorised $x^n - 1$ as $(x+1)(x^{n-1}-x^{n-2}+x^{n-3} - ...+(-1)^n)$. Then i did the reduction, and I'm left with calculating the limit:
      $$lim_{ntoinfty}[frac{1}{n}-frac{1}{n-1}+frac{1}{n-2} -...+(-1)^{n-1}frac{1}{2} + (-1)^n] = lim_{ntoinfty}[sum_{k=0}^{n-1}(-1)^kfrac{1}{n-k}]$$
      I know I can calculate the fraction as a Riemann Sum using the integral of the function $frac{1}{1-x}$, however I don't know how to deal with the changing signs. How should i approach this? Is there some sort of rule or common practice to deal with alternating signs?






      sequences-and-series limits







      share|cite|improve this question











      $endgroup$




      So, I have got the series: $$I_n = int_0^1frac{x^n}{1+x}dx$$ and my task is to find the limit
      $$lim_{ntoinfty}(I_n)$$
      I have added 1 and subtracted 1 from the numerator and factorised $x^n - 1$ as $(x+1)(x^{n-1}-x^{n-2}+x^{n-3} - ...+(-1)^n)$. Then i did the reduction, and I'm left with calculating the limit:
      $$lim_{ntoinfty}[frac{1}{n}-frac{1}{n-1}+frac{1}{n-2} -...+(-1)^{n-1}frac{1}{2} + (-1)^n] = lim_{ntoinfty}[sum_{k=0}^{n-1}(-1)^kfrac{1}{n-k}]$$
      I know I can calculate the fraction as a Riemann Sum using the integral of the function $frac{1}{1-x}$, however I don't know how to deal with the changing signs. How should i approach this? Is there some sort of rule or common practice to deal with alternating signs?






      sequences-and-series limits




      sequences-and-series limits






      share|cite|improve this question















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      edited Jan 25 at 8:36







      Constantin

















      asked Jan 24 at 21:20









      ConstantinConstantin

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          3 Answers
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          $begingroup$

          $$I_nleqint_0^1x^ndx=frac 1{n+1}$$ so $$lim_{nrightarrow +infty}I_n=0$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Just to make sure: you are basing this on the squeeze theorem? 0 <= In <= 1/(n+1) therefore 0 <= lim(In) <= 0 when n goes to infinity?
            $endgroup$
            – Constantin
            Jan 25 at 15:27










          • $begingroup$
            That's right. You got it.
            $endgroup$
            – Stefan Lafon
            Jan 25 at 15:34



















          2












          $begingroup$

          Note that for all $n$ we have :



          $$forall xin [0,1], frac{x^n}{1+x} leq 1$$



          A constant function is clearly integrable, hence we can apply the dominated convergence theorem to get :



          $$lim_{n to infty}int_0^1 frac{x^n}{1+x} mathrm{d}x = int_0^1, lim_{n to infty} frac{x^n}{1+x} mathrm{d}x = 0$$






          share|cite|improve this answer









          $endgroup$





















            1












            $begingroup$

            Proceeding as in the OP, note that we have two cases



            $$frac{1+x^n}{1+x}=begin{cases}sum_{k=1}^{n} (-1)^{k-1}x^{k-1}&,n,,text{odd}\\
            frac2{1+x}-sum_{k=1}^{n}(-1)^{k-1}x^{k-1}&,n,,text{even}end{cases}$$



            Integrating over $[0,1]$, letting $nto infty$, and using the Taylor series representation for $log(1+x)$ for $x=1$ reveals



            $$lim_{ntoinfty}int_0^1 frac{1+x^n}{1+x},dx=log(2) $$



            Hence, we have



            $$lim_{ntoinfty}int_0^1 frac{x^n}{1+x},dx=log(2)-log(2)=0$$




            Rather than proceed as in the OP, we write $frac1{1+x}=sum_{m=0}^{infty}(-x)^m$. Then, we see that



            $$begin{align}
            int_0^1 frac{x^n}{1+x} ,dx&=sum_{m=0}^{infty} (-1)^m int_0^1 x^{n+m},dx\\
            &=sum_{m=n+1}^{infty}(-1)^{m-n-1} frac1{m}tag1
            end{align}$$



            As the series $sum_{m=1}^{infty}frac{(-1)^{m-n-1}}{m}$ converges, the series on the right hand side of $(1)$ approaches $0$ as $nto infty$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy New Year! ;-)
              $endgroup$
              – Mark Viola
              Jan 30 at 5:04













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            3 Answers
            3






            active

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            3 Answers
            3






            active

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            active

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            4












            $begingroup$

            $$I_nleqint_0^1x^ndx=frac 1{n+1}$$ so $$lim_{nrightarrow +infty}I_n=0$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Just to make sure: you are basing this on the squeeze theorem? 0 <= In <= 1/(n+1) therefore 0 <= lim(In) <= 0 when n goes to infinity?
              $endgroup$
              – Constantin
              Jan 25 at 15:27










            • $begingroup$
              That's right. You got it.
              $endgroup$
              – Stefan Lafon
              Jan 25 at 15:34
















            4












            $begingroup$

            $$I_nleqint_0^1x^ndx=frac 1{n+1}$$ so $$lim_{nrightarrow +infty}I_n=0$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Just to make sure: you are basing this on the squeeze theorem? 0 <= In <= 1/(n+1) therefore 0 <= lim(In) <= 0 when n goes to infinity?
              $endgroup$
              – Constantin
              Jan 25 at 15:27










            • $begingroup$
              That's right. You got it.
              $endgroup$
              – Stefan Lafon
              Jan 25 at 15:34














            4












            4








            4





            $begingroup$

            $$I_nleqint_0^1x^ndx=frac 1{n+1}$$ so $$lim_{nrightarrow +infty}I_n=0$$






            share|cite|improve this answer









            $endgroup$



            $$I_nleqint_0^1x^ndx=frac 1{n+1}$$ so $$lim_{nrightarrow +infty}I_n=0$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 24 at 21:29









            Stefan LafonStefan Lafon

            2,83519




            2,83519












            • $begingroup$
              Just to make sure: you are basing this on the squeeze theorem? 0 <= In <= 1/(n+1) therefore 0 <= lim(In) <= 0 when n goes to infinity?
              $endgroup$
              – Constantin
              Jan 25 at 15:27










            • $begingroup$
              That's right. You got it.
              $endgroup$
              – Stefan Lafon
              Jan 25 at 15:34


















            • $begingroup$
              Just to make sure: you are basing this on the squeeze theorem? 0 <= In <= 1/(n+1) therefore 0 <= lim(In) <= 0 when n goes to infinity?
              $endgroup$
              – Constantin
              Jan 25 at 15:27










            • $begingroup$
              That's right. You got it.
              $endgroup$
              – Stefan Lafon
              Jan 25 at 15:34
















            $begingroup$
            Just to make sure: you are basing this on the squeeze theorem? 0 <= In <= 1/(n+1) therefore 0 <= lim(In) <= 0 when n goes to infinity?
            $endgroup$
            – Constantin
            Jan 25 at 15:27




            $begingroup$
            Just to make sure: you are basing this on the squeeze theorem? 0 <= In <= 1/(n+1) therefore 0 <= lim(In) <= 0 when n goes to infinity?
            $endgroup$
            – Constantin
            Jan 25 at 15:27












            $begingroup$
            That's right. You got it.
            $endgroup$
            – Stefan Lafon
            Jan 25 at 15:34




            $begingroup$
            That's right. You got it.
            $endgroup$
            – Stefan Lafon
            Jan 25 at 15:34











            2












            $begingroup$

            Note that for all $n$ we have :



            $$forall xin [0,1], frac{x^n}{1+x} leq 1$$



            A constant function is clearly integrable, hence we can apply the dominated convergence theorem to get :



            $$lim_{n to infty}int_0^1 frac{x^n}{1+x} mathrm{d}x = int_0^1, lim_{n to infty} frac{x^n}{1+x} mathrm{d}x = 0$$






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              Note that for all $n$ we have :



              $$forall xin [0,1], frac{x^n}{1+x} leq 1$$



              A constant function is clearly integrable, hence we can apply the dominated convergence theorem to get :



              $$lim_{n to infty}int_0^1 frac{x^n}{1+x} mathrm{d}x = int_0^1, lim_{n to infty} frac{x^n}{1+x} mathrm{d}x = 0$$






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Note that for all $n$ we have :



                $$forall xin [0,1], frac{x^n}{1+x} leq 1$$



                A constant function is clearly integrable, hence we can apply the dominated convergence theorem to get :



                $$lim_{n to infty}int_0^1 frac{x^n}{1+x} mathrm{d}x = int_0^1, lim_{n to infty} frac{x^n}{1+x} mathrm{d}x = 0$$






                share|cite|improve this answer









                $endgroup$



                Note that for all $n$ we have :



                $$forall xin [0,1], frac{x^n}{1+x} leq 1$$



                A constant function is clearly integrable, hence we can apply the dominated convergence theorem to get :



                $$lim_{n to infty}int_0^1 frac{x^n}{1+x} mathrm{d}x = int_0^1, lim_{n to infty} frac{x^n}{1+x} mathrm{d}x = 0$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 24 at 21:49









                ThinkingThinking

                1,13916




                1,13916























                    1












                    $begingroup$

                    Proceeding as in the OP, note that we have two cases



                    $$frac{1+x^n}{1+x}=begin{cases}sum_{k=1}^{n} (-1)^{k-1}x^{k-1}&,n,,text{odd}\\
                    frac2{1+x}-sum_{k=1}^{n}(-1)^{k-1}x^{k-1}&,n,,text{even}end{cases}$$



                    Integrating over $[0,1]$, letting $nto infty$, and using the Taylor series representation for $log(1+x)$ for $x=1$ reveals



                    $$lim_{ntoinfty}int_0^1 frac{1+x^n}{1+x},dx=log(2) $$



                    Hence, we have



                    $$lim_{ntoinfty}int_0^1 frac{x^n}{1+x},dx=log(2)-log(2)=0$$




                    Rather than proceed as in the OP, we write $frac1{1+x}=sum_{m=0}^{infty}(-x)^m$. Then, we see that



                    $$begin{align}
                    int_0^1 frac{x^n}{1+x} ,dx&=sum_{m=0}^{infty} (-1)^m int_0^1 x^{n+m},dx\\
                    &=sum_{m=n+1}^{infty}(-1)^{m-n-1} frac1{m}tag1
                    end{align}$$



                    As the series $sum_{m=1}^{infty}frac{(-1)^{m-n-1}}{m}$ converges, the series on the right hand side of $(1)$ approaches $0$ as $nto infty$.






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy New Year! ;-)
                      $endgroup$
                      – Mark Viola
                      Jan 30 at 5:04


















                    1












                    $begingroup$

                    Proceeding as in the OP, note that we have two cases



                    $$frac{1+x^n}{1+x}=begin{cases}sum_{k=1}^{n} (-1)^{k-1}x^{k-1}&,n,,text{odd}\\
                    frac2{1+x}-sum_{k=1}^{n}(-1)^{k-1}x^{k-1}&,n,,text{even}end{cases}$$



                    Integrating over $[0,1]$, letting $nto infty$, and using the Taylor series representation for $log(1+x)$ for $x=1$ reveals



                    $$lim_{ntoinfty}int_0^1 frac{1+x^n}{1+x},dx=log(2) $$



                    Hence, we have



                    $$lim_{ntoinfty}int_0^1 frac{x^n}{1+x},dx=log(2)-log(2)=0$$




                    Rather than proceed as in the OP, we write $frac1{1+x}=sum_{m=0}^{infty}(-x)^m$. Then, we see that



                    $$begin{align}
                    int_0^1 frac{x^n}{1+x} ,dx&=sum_{m=0}^{infty} (-1)^m int_0^1 x^{n+m},dx\\
                    &=sum_{m=n+1}^{infty}(-1)^{m-n-1} frac1{m}tag1
                    end{align}$$



                    As the series $sum_{m=1}^{infty}frac{(-1)^{m-n-1}}{m}$ converges, the series on the right hand side of $(1)$ approaches $0$ as $nto infty$.






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy New Year! ;-)
                      $endgroup$
                      – Mark Viola
                      Jan 30 at 5:04
















                    1












                    1








                    1





                    $begingroup$

                    Proceeding as in the OP, note that we have two cases



                    $$frac{1+x^n}{1+x}=begin{cases}sum_{k=1}^{n} (-1)^{k-1}x^{k-1}&,n,,text{odd}\\
                    frac2{1+x}-sum_{k=1}^{n}(-1)^{k-1}x^{k-1}&,n,,text{even}end{cases}$$



                    Integrating over $[0,1]$, letting $nto infty$, and using the Taylor series representation for $log(1+x)$ for $x=1$ reveals



                    $$lim_{ntoinfty}int_0^1 frac{1+x^n}{1+x},dx=log(2) $$



                    Hence, we have



                    $$lim_{ntoinfty}int_0^1 frac{x^n}{1+x},dx=log(2)-log(2)=0$$




                    Rather than proceed as in the OP, we write $frac1{1+x}=sum_{m=0}^{infty}(-x)^m$. Then, we see that



                    $$begin{align}
                    int_0^1 frac{x^n}{1+x} ,dx&=sum_{m=0}^{infty} (-1)^m int_0^1 x^{n+m},dx\\
                    &=sum_{m=n+1}^{infty}(-1)^{m-n-1} frac1{m}tag1
                    end{align}$$



                    As the series $sum_{m=1}^{infty}frac{(-1)^{m-n-1}}{m}$ converges, the series on the right hand side of $(1)$ approaches $0$ as $nto infty$.






                    share|cite|improve this answer











                    $endgroup$



                    Proceeding as in the OP, note that we have two cases



                    $$frac{1+x^n}{1+x}=begin{cases}sum_{k=1}^{n} (-1)^{k-1}x^{k-1}&,n,,text{odd}\\
                    frac2{1+x}-sum_{k=1}^{n}(-1)^{k-1}x^{k-1}&,n,,text{even}end{cases}$$



                    Integrating over $[0,1]$, letting $nto infty$, and using the Taylor series representation for $log(1+x)$ for $x=1$ reveals



                    $$lim_{ntoinfty}int_0^1 frac{1+x^n}{1+x},dx=log(2) $$



                    Hence, we have



                    $$lim_{ntoinfty}int_0^1 frac{x^n}{1+x},dx=log(2)-log(2)=0$$




                    Rather than proceed as in the OP, we write $frac1{1+x}=sum_{m=0}^{infty}(-x)^m$. Then, we see that



                    $$begin{align}
                    int_0^1 frac{x^n}{1+x} ,dx&=sum_{m=0}^{infty} (-1)^m int_0^1 x^{n+m},dx\\
                    &=sum_{m=n+1}^{infty}(-1)^{m-n-1} frac1{m}tag1
                    end{align}$$



                    As the series $sum_{m=1}^{infty}frac{(-1)^{m-n-1}}{m}$ converges, the series on the right hand side of $(1)$ approaches $0$ as $nto infty$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 25 at 1:01

























                    answered Jan 24 at 22:03









                    Mark ViolaMark Viola

                    133k1277176




                    133k1277176












                    • $begingroup$
                      Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy New Year! ;-)
                      $endgroup$
                      – Mark Viola
                      Jan 30 at 5:04




















                    • $begingroup$
                      Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy New Year! ;-)
                      $endgroup$
                      – Mark Viola
                      Jan 30 at 5:04


















                    $begingroup$
                    Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy New Year! ;-)
                    $endgroup$
                    – Mark Viola
                    Jan 30 at 5:04






                    $begingroup$
                    Please let me know how I can improve my answer. I really want to give you the best answer I can. And Happy New Year! ;-)
                    $endgroup$
                    – Mark Viola
                    Jan 30 at 5:04




















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                    Dobbiaco

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                    Dobbiaco comune ( IT ) Dobbiaco ( DE ) Toblach Dobbiaco Vecchia Localizzazione Stato   Italia Regione Trentino-Alto Adige Provincia Bolzano Amministrazione Sindaco Guido Bocher (lista civica Indipendenti-Unabhängige) dal 10-5-2015 (2º mandato) Lingue ufficiali Tedesco, Italiano Territorio Coordinate 46°44′07.35″N 12°13′20.2″E  /  46.735375°N 12.222278°E 46.735375; 12.222278  ( Dobbiaco ) Coordinate: 46°44′07.35″N 12°13′20.2″E  /  46.735375°N 12.222278°E 46.735375; 12.222278  ( Dobbiaco ) Altitudine 1 213 m s.l.m. Superficie 125,42 km² Abitanti 3 356 [2] (31-7-2017) Densità 26,76 ab./km² Frazioni Carbonin ( Schluderbach ), Santa Maria ( Aufkirchen ), Valle San Silvestro ( Wahlen ) Comuni confinanti Auronzo di Cadore (BL), Braies, Cortina d'Ampezzo (BL), Innervillgraten (AT-7), San Candido, Sesto, Valle di Casies, Villabassa Altre informazioni Cod. postale 39034 Prefisso 0474 Fuso...
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