Science:Math Exam Resources/Courses/MATH105/April 2011/Question 09 (c)

MATH105 April 2011

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Question 09 (c)
Each of the short-answer questions below is worth 5 points. Put your answer in the box provided and show your work. No credit will be given for the answer without the correct accompanying work. Express the limit $\lim _{n\rightarrow \infty }{\frac {1}{n}}\sum _{k=1}^{n}\ln \left(1+{\frac {k}{n}}\right)$ as a definite integral, and leave it in the form of an integral. Do not evaluate it!

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
The Riemann sum formula using the right end points for a function $f(x)$ is $f(x)=R_{n}=\sum _{k=1}^{n}f(x_{k})\Delta x$ where $x_{k}=a+k\Delta x\quad \quad \Delta x={\frac {b-a}{n}}$

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Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The Riemann sum formula using the right end points for a function $f(x)$ is $f(x)=R_{n}=\sum _{k=1}^{n}f(x_{k})\Delta x$ where $x_{k}=a+k\Delta x\quad \quad \Delta x={\frac {b-a}{n}}$ In our example, comparing the terms, we see that $\Delta x={\frac {1}{n}}$ and so $b-a=1$ Also, we have that $f(x_{k})=\ln \left(1+{\frac {k}{n}}\right)$ So taking $f(x)=\ln(1+x)$, we have $x_{k}=a+k\Delta x=a+{\frac {k}{n}}={\frac {k}{n}}$ so we have $a=0$ and hence since $b-a=1$ and so $b=b-0=1$. Plugging all this information in, we have $\lim _{n\rightarrow \infty }{\frac {1}{n}}\sum _{k=1}^{n}\ln \left(1+{\frac {k}{n}}\right)=\int _{0}^{1}\ln(1+x)\,dx$

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The Riemann sum formula using the right end points for a function $f(x)$ is $f(x)=R_{n}=\sum _{k=1}^{n}f(x_{k})\Delta x$ where $x_{k}=a+k\Delta x\quad \quad \Delta x={\frac {b-a}{n}}$ In our example, comparing the terms, we see that $\Delta x={\frac {1}{n}}$ and so $b-a=1$ Also, we have that $f(x_{k})=\ln \left(1+{\frac {k}{n}}\right)$ So taking $f(x)=\ln(x)$, we have $x_{k}=a+k\Delta x=a+{\frac {k}{n}}=1+{\frac {k}{n}}$ so we have $a=1$ and hence since $b-a=1$ and so $b-1=1$ giving $b=2$. Plugging all this information in, we have $\lim _{n\rightarrow \infty }{\frac {1}{n}}\sum _{k=1}^{n}\ln \left(1+{\frac {k}{n}}\right)=\int _{1}^{2}\ln(x)\,dx$

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