Integral Method to find upper and lower bounds on the sum












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$begingroup$


I got this problem wrong on my exam and I know why I got it wrong but I am not sure how to find the right answer.



This is the question:
Consider$sum_{k=1}^n (3+ 2/k^3)$. Use the integral method to obtain upper and lower bound on the sum. Do not split the summation into $sum_{k=1}^n 3+sum_{k=1}^n ( 2/k^3)$



This is the formula we had to use: Click here to see the approximation integral formula



This function is decreasing and I was able to calculate the lower bound, but I am not sure how to do the upper bound becuase this is what I get $ int_{0}^{n}$(3+2/k^3)dx and after taking the antiderivative. I get 3k - (1/k^2) from 0 to n. so if I take the integral from 0 to n then 1/k^2 becomes undefined. I tried to change the integral from -1 to n but I got it wrong on the exam.



Could someone please tell me how to solve this? Thanks






definite-integrals approximate-integration







share|cite|improve this question









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    0












    $begingroup$


    I got this problem wrong on my exam and I know why I got it wrong but I am not sure how to find the right answer.



    This is the question:
    Consider$sum_{k=1}^n (3+ 2/k^3)$. Use the integral method to obtain upper and lower bound on the sum. Do not split the summation into $sum_{k=1}^n 3+sum_{k=1}^n ( 2/k^3)$



    This is the formula we had to use: Click here to see the approximation integral formula



    This function is decreasing and I was able to calculate the lower bound, but I am not sure how to do the upper bound becuase this is what I get $ int_{0}^{n}$(3+2/k^3)dx and after taking the antiderivative. I get 3k - (1/k^2) from 0 to n. so if I take the integral from 0 to n then 1/k^2 becomes undefined. I tried to change the integral from -1 to n but I got it wrong on the exam.



    Could someone please tell me how to solve this? Thanks






    definite-integrals approximate-integration







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I got this problem wrong on my exam and I know why I got it wrong but I am not sure how to find the right answer.



      This is the question:
      Consider$sum_{k=1}^n (3+ 2/k^3)$. Use the integral method to obtain upper and lower bound on the sum. Do not split the summation into $sum_{k=1}^n 3+sum_{k=1}^n ( 2/k^3)$



      This is the formula we had to use: Click here to see the approximation integral formula



      This function is decreasing and I was able to calculate the lower bound, but I am not sure how to do the upper bound becuase this is what I get $ int_{0}^{n}$(3+2/k^3)dx and after taking the antiderivative. I get 3k - (1/k^2) from 0 to n. so if I take the integral from 0 to n then 1/k^2 becomes undefined. I tried to change the integral from -1 to n but I got it wrong on the exam.



      Could someone please tell me how to solve this? Thanks






      definite-integrals approximate-integration







      share|cite|improve this question









      $endgroup$




      I got this problem wrong on my exam and I know why I got it wrong but I am not sure how to find the right answer.



      This is the question:
      Consider$sum_{k=1}^n (3+ 2/k^3)$. Use the integral method to obtain upper and lower bound on the sum. Do not split the summation into $sum_{k=1}^n 3+sum_{k=1}^n ( 2/k^3)$



      This is the formula we had to use: Click here to see the approximation integral formula



      This function is decreasing and I was able to calculate the lower bound, but I am not sure how to do the upper bound becuase this is what I get $ int_{0}^{n}$(3+2/k^3)dx and after taking the antiderivative. I get 3k - (1/k^2) from 0 to n. so if I take the integral from 0 to n then 1/k^2 becomes undefined. I tried to change the integral from -1 to n but I got it wrong on the exam.



      Could someone please tell me how to solve this? Thanks






      definite-integrals approximate-integration




      definite-integrals approximate-integration






      share|cite|improve this question













      share|cite|improve this question











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      share|cite|improve this question










      asked Apr 2 '17 at 9:35









      jenkins342jenkins342

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          1 Answer
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          $begingroup$

          Hint. You may write
          $$
          begin{align}
          sum_{k=1}^nleft(3+frac1{k^3} right)&=left(3+frac1{1^3} right)+sum_{k=color{red}{2}}^nleft(3+frac1{k^3} right)
          \\&le left(3+frac1{1^3}right)+int_{color{red}{2-1}}^nleft(3+frac1{x^3} right)dx
          end{align}
          $$ then getting easily an upper bound.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thanks! Will I need to calculate the derivative of 5 in the integration part or just take the derivative of what's inside the summation and then multiply by 5? Also, for the lower bound, I can use the original summation without changing the bounds?
            $endgroup$
            – jenkins342
            Apr 2 '17 at 10:07













          Your Answer





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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          0












          $begingroup$

          Hint. You may write
          $$
          begin{align}
          sum_{k=1}^nleft(3+frac1{k^3} right)&=left(3+frac1{1^3} right)+sum_{k=color{red}{2}}^nleft(3+frac1{k^3} right)
          \\&le left(3+frac1{1^3}right)+int_{color{red}{2-1}}^nleft(3+frac1{x^3} right)dx
          end{align}
          $$ then getting easily an upper bound.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thanks! Will I need to calculate the derivative of 5 in the integration part or just take the derivative of what's inside the summation and then multiply by 5? Also, for the lower bound, I can use the original summation without changing the bounds?
            $endgroup$
            – jenkins342
            Apr 2 '17 at 10:07


















          0












          $begingroup$

          Hint. You may write
          $$
          begin{align}
          sum_{k=1}^nleft(3+frac1{k^3} right)&=left(3+frac1{1^3} right)+sum_{k=color{red}{2}}^nleft(3+frac1{k^3} right)
          \\&le left(3+frac1{1^3}right)+int_{color{red}{2-1}}^nleft(3+frac1{x^3} right)dx
          end{align}
          $$ then getting easily an upper bound.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            thanks! Will I need to calculate the derivative of 5 in the integration part or just take the derivative of what's inside the summation and then multiply by 5? Also, for the lower bound, I can use the original summation without changing the bounds?
            $endgroup$
            – jenkins342
            Apr 2 '17 at 10:07
















          0












          0








          0





          $begingroup$

          Hint. You may write
          $$
          begin{align}
          sum_{k=1}^nleft(3+frac1{k^3} right)&=left(3+frac1{1^3} right)+sum_{k=color{red}{2}}^nleft(3+frac1{k^3} right)
          \\&le left(3+frac1{1^3}right)+int_{color{red}{2-1}}^nleft(3+frac1{x^3} right)dx
          end{align}
          $$ then getting easily an upper bound.






          share|cite|improve this answer









          $endgroup$



          Hint. You may write
          $$
          begin{align}
          sum_{k=1}^nleft(3+frac1{k^3} right)&=left(3+frac1{1^3} right)+sum_{k=color{red}{2}}^nleft(3+frac1{k^3} right)
          \\&le left(3+frac1{1^3}right)+int_{color{red}{2-1}}^nleft(3+frac1{x^3} right)dx
          end{align}
          $$ then getting easily an upper bound.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Apr 2 '17 at 9:45









          Olivier OloaOlivier Oloa

          109k17178294




          109k17178294












          • $begingroup$
            thanks! Will I need to calculate the derivative of 5 in the integration part or just take the derivative of what's inside the summation and then multiply by 5? Also, for the lower bound, I can use the original summation without changing the bounds?
            $endgroup$
            – jenkins342
            Apr 2 '17 at 10:07




















          • $begingroup$
            thanks! Will I need to calculate the derivative of 5 in the integration part or just take the derivative of what's inside the summation and then multiply by 5? Also, for the lower bound, I can use the original summation without changing the bounds?
            $endgroup$
            – jenkins342
            Apr 2 '17 at 10:07


















          $begingroup$
          thanks! Will I need to calculate the derivative of 5 in the integration part or just take the derivative of what's inside the summation and then multiply by 5? Also, for the lower bound, I can use the original summation without changing the bounds?
          $endgroup$
          – jenkins342
          Apr 2 '17 at 10:07






          $begingroup$
          thanks! Will I need to calculate the derivative of 5 in the integration part or just take the derivative of what's inside the summation and then multiply by 5? Also, for the lower bound, I can use the original summation without changing the bounds?
          $endgroup$
          – jenkins342
          Apr 2 '17 at 10:07




















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