Science:Math Exam Resources/Courses/MATH101/April 2006/Question 07

MATH101 April 2006

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q1 (e)  •  Q1 (f)  •  Q1 (g)  •  Q1 (h)  •  Q1 (i)  •  Q1 (j)  •  Q1 (k)  •  Q2 (a)  •  Q2 (b)  •  Q2 (c)  •  Q2 (d)  •  Q3 (a)  •  Q3 (b)  •  Q3 (c)  •  Q3 (d)  •  Q4 (a)  •  Q4 (b)  •  Q5  •  Q6  •  Q7  •  

   Other MATH101 Exams

  •  April 2011  •  April 2010  •  April 2009  •  April 2008  •  April 2007  •  April 2005  •  April 2006  •  April 2012  •  April 2013  •  April 2014  •  April 2015  •  April 2016  •  April 2018  •  April 2017  •  

Question 07
Full-Solution Problems. Justify your answers and show all your work. Simplification of answers is not required. Evaluate $\lim _{n\to \infty }\sum _{i=1}^{n}{\frac {n}{(i+n)(i+2n)}}$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
The notation should be awfully reminiscent of Riemann sums. However, the way the sum is presented makes it look a bit mysterious from the Riemann sums we are used to. What or where is the $\Delta x$ value? What or where is the $x_{i}$ value?

Hint 2
Try factoring out an n from each of the two factors in the denominator.

Hint 3
The values should look like $\displaystyle \Delta x={\frac {1}{n}}$ $\displaystyle x_{i}=1+{\frac {i}{n}}$ $\displaystyle a=1$ and so with the above, we can deduce that $\displaystyle b=2$

Hint 4
Convert to an integral and now use partial fractions to take you home.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We proceed as in the hint (specifically hint 3). Notice that $\lim _{n\to \infty }\sum _{i=1}^{n}{\frac {n}{(i+n)(i+2n)}}=\lim _{n\to \infty }\sum _{i=1}^{n}{\frac {1}{n}}\cdot {\frac {1}{({\tfrac {i}{n}}+1)({\tfrac {i}{n}}+2)}}$ From this, it is clear that $\Delta x={\tfrac {1}{n}}$ and that we should probably set $x_{i}=1+{\tfrac {i}{n}}$ and so $a=1$ and thus, combining this with the $\Delta x$ information, we have ${\frac {1}{n}}=\Delta x={\frac {b-a}{n}}={\frac {b-1}{n}}$ and solving for $b$ gives $b=2$. Hence, we have $\lim _{n\to \infty }\sum _{i=1}^{n}{\frac {n}{(i+n)(i+2n)}}=\int _{1}^{2}{\frac {dx}{x(x+1)}}$ We've reduced the problem to solving an integral. This integral is ripe for the partial fraction algorithm so let's apply that. Let ${\frac {1}{x(x+1)}}={\frac {A}{x}}+{\frac {B}{x+1}}={\frac {A(x+1)+Bx}{x(x+1)}}$ and this gives rise to $\displaystyle 1=A(x+1)+Bx$ Plugging in $x=0$ in the above yields $\displaystyle 1=A((0)+1)+B(0)=A$ and so $A=1$. Plugging in $x=-1$ into the equation gives $\displaystyle 1=A((-1)+1)+B(-1)=-B$ and so $B=-1$. Thus, we have ${\begin{aligned}\lim _{n\to \infty }\sum _{i=1}^{n}{\frac {n}{(i+n)(i+2n)}}&=\int _{1}^{2}{\frac {dx}{x(x+1)}}\\&=\int _{1}^{2}{\frac {dx}{x}}-\int _{1}^{2}{\frac {dx}{x+1}}\\&=\left.\ln |x|\right|_{1}^{2}-\left.\ln |x+1|\right|_{1}^{2}\\&=\ln(2)-\ln(1)-(\ln(3)-\ln(2))\\&=2\ln(2)-\ln(3)\end{aligned}}$ completing the question.

Click here for similar questions

MER QGH flag, MER QGQ flag, MER QGS flag, MER RT flag, MER Tag Partial fractions, MER Tag Riemann sum, Pages using DynamicPageList3 parser function, Pages using DynamicPageList3 parser tag