Find the limit $lim_{ntoinfty}frac{x_n + x_n^2 + cdots + x_n^k - k}{x_n - 1}$ given $x_n ne 1$ and $lim x_n =...












2












$begingroup$



Let $x_n$ denote a sequence, $ninBbb N$. Evaluate the limit:
$$
lim_{ntoinfty} frac{x_n + x_n^2 + cdots + x_n^k - k}{x_n - 1}, kinBbb N
$$
given
$$
lim_{ntoinfty} x_n = 1 \
x_n ne 1
$$




I'm interested in verifying the following results. Denote:
$$
y_n = frac{x_n + x_n^2 + cdots + x_n^k - k}{x_n - 1}
$$
After some trials long division seem to produce a pattern here. Not sure how to put it here in a fancy way, so I will post the result only. It appears that:
$$
y_n = x_n^{k-1} + 2x_n^{k-2}+cdots + (k-1)x_n + k
$$



It is given that $lim x_n = 1$ therefore we may use the following properties:
$$
lim(a_n + b_n) = lim a_n + lim b_n\
lim(a_n cdot b_n) = lim a_n cdot lim b_n
$$



In particular:
$$
y_n = x_n^{k-1} + 2x_n^{k-2}+cdots + (k-1)x_n + k\
lim_{ntoinfty}y_n =lim_{ntoinfty}left(x_n^{k-1} + 2x_n^{k-2}+cdots + (k-1)x_n + kright)
$$



Then since $lim x_n = 1$ and by the sum of first $k$ integers:
$$
lim_{ntoinfty} y_n = frac{k(k+1)}{2}
$$



Could you please verify whether the reasoning above is correct or not and point to the mistakes in case of any? Thank you!






calculus sequences-and-series limits proof-verification







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$endgroup$












  • $begingroup$
    Your argument is correct. The limit is indeed $frac{1}{2}k(k+1)$
    $endgroup$
    – Crostul
    Jan 15 at 16:05










  • $begingroup$
    Not to dampen your spirit, but perhaps a proof using L'Hopital would be shorter?
    $endgroup$
    – Keen-ameteur
    Jan 15 at 16:11










  • $begingroup$
    @Keen-ameteur Unfortunately I'm not supposed to use derivatives or even Stolz-Cesaro since none of them has been yet defined at where i am now in the book.
    $endgroup$
    – roman
    Jan 15 at 16:13










  • $begingroup$
    Okay, then nevermind.
    $endgroup$
    – Keen-ameteur
    Jan 15 at 16:14
















2












$begingroup$



Let $x_n$ denote a sequence, $ninBbb N$. Evaluate the limit:
$$
lim_{ntoinfty} frac{x_n + x_n^2 + cdots + x_n^k - k}{x_n - 1}, kinBbb N
$$
given
$$
lim_{ntoinfty} x_n = 1 \
x_n ne 1
$$




I'm interested in verifying the following results. Denote:
$$
y_n = frac{x_n + x_n^2 + cdots + x_n^k - k}{x_n - 1}
$$
After some trials long division seem to produce a pattern here. Not sure how to put it here in a fancy way, so I will post the result only. It appears that:
$$
y_n = x_n^{k-1} + 2x_n^{k-2}+cdots + (k-1)x_n + k
$$



It is given that $lim x_n = 1$ therefore we may use the following properties:
$$
lim(a_n + b_n) = lim a_n + lim b_n\
lim(a_n cdot b_n) = lim a_n cdot lim b_n
$$



In particular:
$$
y_n = x_n^{k-1} + 2x_n^{k-2}+cdots + (k-1)x_n + k\
lim_{ntoinfty}y_n =lim_{ntoinfty}left(x_n^{k-1} + 2x_n^{k-2}+cdots + (k-1)x_n + kright)
$$



Then since $lim x_n = 1$ and by the sum of first $k$ integers:
$$
lim_{ntoinfty} y_n = frac{k(k+1)}{2}
$$



Could you please verify whether the reasoning above is correct or not and point to the mistakes in case of any? Thank you!






calculus sequences-and-series limits proof-verification







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your argument is correct. The limit is indeed $frac{1}{2}k(k+1)$
    $endgroup$
    – Crostul
    Jan 15 at 16:05










  • $begingroup$
    Not to dampen your spirit, but perhaps a proof using L'Hopital would be shorter?
    $endgroup$
    – Keen-ameteur
    Jan 15 at 16:11










  • $begingroup$
    @Keen-ameteur Unfortunately I'm not supposed to use derivatives or even Stolz-Cesaro since none of them has been yet defined at where i am now in the book.
    $endgroup$
    – roman
    Jan 15 at 16:13










  • $begingroup$
    Okay, then nevermind.
    $endgroup$
    – Keen-ameteur
    Jan 15 at 16:14














2












2








2


1



$begingroup$



Let $x_n$ denote a sequence, $ninBbb N$. Evaluate the limit:
$$
lim_{ntoinfty} frac{x_n + x_n^2 + cdots + x_n^k - k}{x_n - 1}, kinBbb N
$$
given
$$
lim_{ntoinfty} x_n = 1 \
x_n ne 1
$$




I'm interested in verifying the following results. Denote:
$$
y_n = frac{x_n + x_n^2 + cdots + x_n^k - k}{x_n - 1}
$$
After some trials long division seem to produce a pattern here. Not sure how to put it here in a fancy way, so I will post the result only. It appears that:
$$
y_n = x_n^{k-1} + 2x_n^{k-2}+cdots + (k-1)x_n + k
$$



It is given that $lim x_n = 1$ therefore we may use the following properties:
$$
lim(a_n + b_n) = lim a_n + lim b_n\
lim(a_n cdot b_n) = lim a_n cdot lim b_n
$$



In particular:
$$
y_n = x_n^{k-1} + 2x_n^{k-2}+cdots + (k-1)x_n + k\
lim_{ntoinfty}y_n =lim_{ntoinfty}left(x_n^{k-1} + 2x_n^{k-2}+cdots + (k-1)x_n + kright)
$$



Then since $lim x_n = 1$ and by the sum of first $k$ integers:
$$
lim_{ntoinfty} y_n = frac{k(k+1)}{2}
$$



Could you please verify whether the reasoning above is correct or not and point to the mistakes in case of any? Thank you!






calculus sequences-and-series limits proof-verification







share|cite|improve this question











$endgroup$





Let $x_n$ denote a sequence, $ninBbb N$. Evaluate the limit:
$$
lim_{ntoinfty} frac{x_n + x_n^2 + cdots + x_n^k - k}{x_n - 1}, kinBbb N
$$
given
$$
lim_{ntoinfty} x_n = 1 \
x_n ne 1
$$




I'm interested in verifying the following results. Denote:
$$
y_n = frac{x_n + x_n^2 + cdots + x_n^k - k}{x_n - 1}
$$
After some trials long division seem to produce a pattern here. Not sure how to put it here in a fancy way, so I will post the result only. It appears that:
$$
y_n = x_n^{k-1} + 2x_n^{k-2}+cdots + (k-1)x_n + k
$$



It is given that $lim x_n = 1$ therefore we may use the following properties:
$$
lim(a_n + b_n) = lim a_n + lim b_n\
lim(a_n cdot b_n) = lim a_n cdot lim b_n
$$



In particular:
$$
y_n = x_n^{k-1} + 2x_n^{k-2}+cdots + (k-1)x_n + k\
lim_{ntoinfty}y_n =lim_{ntoinfty}left(x_n^{k-1} + 2x_n^{k-2}+cdots + (k-1)x_n + kright)
$$



Then since $lim x_n = 1$ and by the sum of first $k$ integers:
$$
lim_{ntoinfty} y_n = frac{k(k+1)}{2}
$$



Could you please verify whether the reasoning above is correct or not and point to the mistakes in case of any? Thank you!






calculus sequences-and-series limits proof-verification




calculus sequences-and-series limits proof-verification






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share|cite|improve this question













share|cite|improve this question




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edited Jan 15 at 15:56







roman

















asked Jan 15 at 15:50









romanroman

2,17321224




2,17321224












  • $begingroup$
    Your argument is correct. The limit is indeed $frac{1}{2}k(k+1)$
    $endgroup$
    – Crostul
    Jan 15 at 16:05










  • $begingroup$
    Not to dampen your spirit, but perhaps a proof using L'Hopital would be shorter?
    $endgroup$
    – Keen-ameteur
    Jan 15 at 16:11










  • $begingroup$
    @Keen-ameteur Unfortunately I'm not supposed to use derivatives or even Stolz-Cesaro since none of them has been yet defined at where i am now in the book.
    $endgroup$
    – roman
    Jan 15 at 16:13










  • $begingroup$
    Okay, then nevermind.
    $endgroup$
    – Keen-ameteur
    Jan 15 at 16:14


















  • $begingroup$
    Your argument is correct. The limit is indeed $frac{1}{2}k(k+1)$
    $endgroup$
    – Crostul
    Jan 15 at 16:05










  • $begingroup$
    Not to dampen your spirit, but perhaps a proof using L'Hopital would be shorter?
    $endgroup$
    – Keen-ameteur
    Jan 15 at 16:11










  • $begingroup$
    @Keen-ameteur Unfortunately I'm not supposed to use derivatives or even Stolz-Cesaro since none of them has been yet defined at where i am now in the book.
    $endgroup$
    – roman
    Jan 15 at 16:13










  • $begingroup$
    Okay, then nevermind.
    $endgroup$
    – Keen-ameteur
    Jan 15 at 16:14
















$begingroup$
Your argument is correct. The limit is indeed $frac{1}{2}k(k+1)$
$endgroup$
– Crostul
Jan 15 at 16:05




$begingroup$
Your argument is correct. The limit is indeed $frac{1}{2}k(k+1)$
$endgroup$
– Crostul
Jan 15 at 16:05












$begingroup$
Not to dampen your spirit, but perhaps a proof using L'Hopital would be shorter?
$endgroup$
– Keen-ameteur
Jan 15 at 16:11




$begingroup$
Not to dampen your spirit, but perhaps a proof using L'Hopital would be shorter?
$endgroup$
– Keen-ameteur
Jan 15 at 16:11












$begingroup$
@Keen-ameteur Unfortunately I'm not supposed to use derivatives or even Stolz-Cesaro since none of them has been yet defined at where i am now in the book.
$endgroup$
– roman
Jan 15 at 16:13




$begingroup$
@Keen-ameteur Unfortunately I'm not supposed to use derivatives or even Stolz-Cesaro since none of them has been yet defined at where i am now in the book.
$endgroup$
– roman
Jan 15 at 16:13












$begingroup$
Okay, then nevermind.
$endgroup$
– Keen-ameteur
Jan 15 at 16:14




$begingroup$
Okay, then nevermind.
$endgroup$
– Keen-ameteur
Jan 15 at 16:14










3 Answers
3






active

oldest

votes


















2












$begingroup$

Rewrite the limit as



$$ sum_{j=1}^k lim_{n to infty} {x_n^j - 1 over x_n - 1} $$



(note that this is a sum of finitely many terms).



Now we can do the division to get



$$ sum_{j=1}^k lim_{n to infty} (1 + x_n + x_n^2 + cdots + x_n^{j-1}) $$



and the limit can be written as a sum of (again, finitely many!) limits. This gives



$$ sum_{j=1}^k sum_{i=0}^{j-1} lim_{n to infty} x_n^i $$



Finally the innermost limit is, for each $n$ and $i$, equal to $(lim_{n to infty} x_n)^i = 1^i = 1$ so this is just



$$ sum_{j=1}^k sum_{i=0}^{j-1} 1 = sum_{j=1}^k j = {k(k+1) over 2}$$



as desired.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    That one is a nice approach
    $endgroup$
    – roman
    Jan 15 at 16:16



















1












$begingroup$

Let $y_n=x_n-1$ and then $lim_{ntoinfty}y_n=1$. Note
begin{eqnarray*}
&&frac{x_n + x_n^2 + cdots + x_n^k - k}{x_n-1}\
&=&frac{(y_n+1)^{k+1}-(k+1)(y_n+1)+k}{y_n^2}\
&=&frac{sum_{i=0}^{k+1}binom{k+1}{i}y_n^i-(k+1)y_n-1}{y_n^2}\
&=&frac{sum_{i=2}^{k+1}binom{k+1}{i}y_n^i-(k+1)y_n}{y_n^2}\
&=&sum_{i=2}^{k+1}binom{k+1}{i}y_n^{i-2}\
end{eqnarray*}
and hence
begin{eqnarray*}
&&lim_{ntoinfty} frac{x_n + x_n^2 + cdots + x_n^k - k}{x_n - 1}\
&=&lim_{ntoinfty}sum_{i=2}^{k+1}binom{k+1}{i}y_n^{i-2}\
&=&binom{k+1}{2}=frac12k(k+1).
end{eqnarray*}






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    This could be realized as a derivative. Let us define
    $$f(x)=x^1+x^2+cdots+x^k,.$$
    Then
    $$lim_{xrightarrow 1}frac{f(x)-f(1)}{x-1}=f'(1)=1+2cdot 1^1+cdots+kcdot 1^{k-1}=frac{k(k+1)}{2},.$$






    share|cite|improve this answer









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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      Rewrite the limit as



      $$ sum_{j=1}^k lim_{n to infty} {x_n^j - 1 over x_n - 1} $$



      (note that this is a sum of finitely many terms).



      Now we can do the division to get



      $$ sum_{j=1}^k lim_{n to infty} (1 + x_n + x_n^2 + cdots + x_n^{j-1}) $$



      and the limit can be written as a sum of (again, finitely many!) limits. This gives



      $$ sum_{j=1}^k sum_{i=0}^{j-1} lim_{n to infty} x_n^i $$



      Finally the innermost limit is, for each $n$ and $i$, equal to $(lim_{n to infty} x_n)^i = 1^i = 1$ so this is just



      $$ sum_{j=1}^k sum_{i=0}^{j-1} 1 = sum_{j=1}^k j = {k(k+1) over 2}$$



      as desired.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        That one is a nice approach
        $endgroup$
        – roman
        Jan 15 at 16:16
















      2












      $begingroup$

      Rewrite the limit as



      $$ sum_{j=1}^k lim_{n to infty} {x_n^j - 1 over x_n - 1} $$



      (note that this is a sum of finitely many terms).



      Now we can do the division to get



      $$ sum_{j=1}^k lim_{n to infty} (1 + x_n + x_n^2 + cdots + x_n^{j-1}) $$



      and the limit can be written as a sum of (again, finitely many!) limits. This gives



      $$ sum_{j=1}^k sum_{i=0}^{j-1} lim_{n to infty} x_n^i $$



      Finally the innermost limit is, for each $n$ and $i$, equal to $(lim_{n to infty} x_n)^i = 1^i = 1$ so this is just



      $$ sum_{j=1}^k sum_{i=0}^{j-1} 1 = sum_{j=1}^k j = {k(k+1) over 2}$$



      as desired.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        That one is a nice approach
        $endgroup$
        – roman
        Jan 15 at 16:16














      2












      2








      2





      $begingroup$

      Rewrite the limit as



      $$ sum_{j=1}^k lim_{n to infty} {x_n^j - 1 over x_n - 1} $$



      (note that this is a sum of finitely many terms).



      Now we can do the division to get



      $$ sum_{j=1}^k lim_{n to infty} (1 + x_n + x_n^2 + cdots + x_n^{j-1}) $$



      and the limit can be written as a sum of (again, finitely many!) limits. This gives



      $$ sum_{j=1}^k sum_{i=0}^{j-1} lim_{n to infty} x_n^i $$



      Finally the innermost limit is, for each $n$ and $i$, equal to $(lim_{n to infty} x_n)^i = 1^i = 1$ so this is just



      $$ sum_{j=1}^k sum_{i=0}^{j-1} 1 = sum_{j=1}^k j = {k(k+1) over 2}$$



      as desired.






      share|cite|improve this answer









      $endgroup$



      Rewrite the limit as



      $$ sum_{j=1}^k lim_{n to infty} {x_n^j - 1 over x_n - 1} $$



      (note that this is a sum of finitely many terms).



      Now we can do the division to get



      $$ sum_{j=1}^k lim_{n to infty} (1 + x_n + x_n^2 + cdots + x_n^{j-1}) $$



      and the limit can be written as a sum of (again, finitely many!) limits. This gives



      $$ sum_{j=1}^k sum_{i=0}^{j-1} lim_{n to infty} x_n^i $$



      Finally the innermost limit is, for each $n$ and $i$, equal to $(lim_{n to infty} x_n)^i = 1^i = 1$ so this is just



      $$ sum_{j=1}^k sum_{i=0}^{j-1} 1 = sum_{j=1}^k j = {k(k+1) over 2}$$



      as desired.







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 15 at 16:13









      Michael LugoMichael Lugo

      18k33576




      18k33576












      • $begingroup$
        That one is a nice approach
        $endgroup$
        – roman
        Jan 15 at 16:16


















      • $begingroup$
        That one is a nice approach
        $endgroup$
        – roman
        Jan 15 at 16:16
















      $begingroup$
      That one is a nice approach
      $endgroup$
      – roman
      Jan 15 at 16:16




      $begingroup$
      That one is a nice approach
      $endgroup$
      – roman
      Jan 15 at 16:16











      1












      $begingroup$

      Let $y_n=x_n-1$ and then $lim_{ntoinfty}y_n=1$. Note
      begin{eqnarray*}
      &&frac{x_n + x_n^2 + cdots + x_n^k - k}{x_n-1}\
      &=&frac{(y_n+1)^{k+1}-(k+1)(y_n+1)+k}{y_n^2}\
      &=&frac{sum_{i=0}^{k+1}binom{k+1}{i}y_n^i-(k+1)y_n-1}{y_n^2}\
      &=&frac{sum_{i=2}^{k+1}binom{k+1}{i}y_n^i-(k+1)y_n}{y_n^2}\
      &=&sum_{i=2}^{k+1}binom{k+1}{i}y_n^{i-2}\
      end{eqnarray*}
      and hence
      begin{eqnarray*}
      &&lim_{ntoinfty} frac{x_n + x_n^2 + cdots + x_n^k - k}{x_n - 1}\
      &=&lim_{ntoinfty}sum_{i=2}^{k+1}binom{k+1}{i}y_n^{i-2}\
      &=&binom{k+1}{2}=frac12k(k+1).
      end{eqnarray*}






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Let $y_n=x_n-1$ and then $lim_{ntoinfty}y_n=1$. Note
        begin{eqnarray*}
        &&frac{x_n + x_n^2 + cdots + x_n^k - k}{x_n-1}\
        &=&frac{(y_n+1)^{k+1}-(k+1)(y_n+1)+k}{y_n^2}\
        &=&frac{sum_{i=0}^{k+1}binom{k+1}{i}y_n^i-(k+1)y_n-1}{y_n^2}\
        &=&frac{sum_{i=2}^{k+1}binom{k+1}{i}y_n^i-(k+1)y_n}{y_n^2}\
        &=&sum_{i=2}^{k+1}binom{k+1}{i}y_n^{i-2}\
        end{eqnarray*}
        and hence
        begin{eqnarray*}
        &&lim_{ntoinfty} frac{x_n + x_n^2 + cdots + x_n^k - k}{x_n - 1}\
        &=&lim_{ntoinfty}sum_{i=2}^{k+1}binom{k+1}{i}y_n^{i-2}\
        &=&binom{k+1}{2}=frac12k(k+1).
        end{eqnarray*}






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Let $y_n=x_n-1$ and then $lim_{ntoinfty}y_n=1$. Note
          begin{eqnarray*}
          &&frac{x_n + x_n^2 + cdots + x_n^k - k}{x_n-1}\
          &=&frac{(y_n+1)^{k+1}-(k+1)(y_n+1)+k}{y_n^2}\
          &=&frac{sum_{i=0}^{k+1}binom{k+1}{i}y_n^i-(k+1)y_n-1}{y_n^2}\
          &=&frac{sum_{i=2}^{k+1}binom{k+1}{i}y_n^i-(k+1)y_n}{y_n^2}\
          &=&sum_{i=2}^{k+1}binom{k+1}{i}y_n^{i-2}\
          end{eqnarray*}
          and hence
          begin{eqnarray*}
          &&lim_{ntoinfty} frac{x_n + x_n^2 + cdots + x_n^k - k}{x_n - 1}\
          &=&lim_{ntoinfty}sum_{i=2}^{k+1}binom{k+1}{i}y_n^{i-2}\
          &=&binom{k+1}{2}=frac12k(k+1).
          end{eqnarray*}






          share|cite|improve this answer









          $endgroup$



          Let $y_n=x_n-1$ and then $lim_{ntoinfty}y_n=1$. Note
          begin{eqnarray*}
          &&frac{x_n + x_n^2 + cdots + x_n^k - k}{x_n-1}\
          &=&frac{(y_n+1)^{k+1}-(k+1)(y_n+1)+k}{y_n^2}\
          &=&frac{sum_{i=0}^{k+1}binom{k+1}{i}y_n^i-(k+1)y_n-1}{y_n^2}\
          &=&frac{sum_{i=2}^{k+1}binom{k+1}{i}y_n^i-(k+1)y_n}{y_n^2}\
          &=&sum_{i=2}^{k+1}binom{k+1}{i}y_n^{i-2}\
          end{eqnarray*}
          and hence
          begin{eqnarray*}
          &&lim_{ntoinfty} frac{x_n + x_n^2 + cdots + x_n^k - k}{x_n - 1}\
          &=&lim_{ntoinfty}sum_{i=2}^{k+1}binom{k+1}{i}y_n^{i-2}\
          &=&binom{k+1}{2}=frac12k(k+1).
          end{eqnarray*}







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          answered Jan 15 at 16:11









          xpaulxpaul

          22.7k24455




          22.7k24455























              1












              $begingroup$

              This could be realized as a derivative. Let us define
              $$f(x)=x^1+x^2+cdots+x^k,.$$
              Then
              $$lim_{xrightarrow 1}frac{f(x)-f(1)}{x-1}=f'(1)=1+2cdot 1^1+cdots+kcdot 1^{k-1}=frac{k(k+1)}{2},.$$






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                This could be realized as a derivative. Let us define
                $$f(x)=x^1+x^2+cdots+x^k,.$$
                Then
                $$lim_{xrightarrow 1}frac{f(x)-f(1)}{x-1}=f'(1)=1+2cdot 1^1+cdots+kcdot 1^{k-1}=frac{k(k+1)}{2},.$$






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  This could be realized as a derivative. Let us define
                  $$f(x)=x^1+x^2+cdots+x^k,.$$
                  Then
                  $$lim_{xrightarrow 1}frac{f(x)-f(1)}{x-1}=f'(1)=1+2cdot 1^1+cdots+kcdot 1^{k-1}=frac{k(k+1)}{2},.$$






                  share|cite|improve this answer









                  $endgroup$



                  This could be realized as a derivative. Let us define
                  $$f(x)=x^1+x^2+cdots+x^k,.$$
                  Then
                  $$lim_{xrightarrow 1}frac{f(x)-f(1)}{x-1}=f'(1)=1+2cdot 1^1+cdots+kcdot 1^{k-1}=frac{k(k+1)}{2},.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 16 at 1:20









                  Robert WolfeRobert Wolfe

                  5,79222763




                  5,79222763






























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