# Evaluate the limit by recognizing the Riemann sum

• August 11th 2009, 08:46 AM
manny
[Solved] Evaluate the limit by recognizing the Riemann sum
Evaluate $\lim_{n\to\infty}\frac{1+2+3+...+n}{n^2+n}$ by first recognizing the sum as a Riemann sum over a partition of [0,1] and then evaluating the corresponding integral.
A. 0
B. 1/2
C. 1/4
D. Not enough information
E. None of the above
----------------------------------------------------------------------------------------------------------------

What is confusing me about the question is whether I'm being asked to compute a Riemann sum or being asked to "reverse engineer" the given function into a Riemann sum. I tend to lean towards the latter so this is what I've done so far...

Given:
$\displaystyle\int^b_a f\left(x\right)\,dx = \lim_{n\to\infty}\displaystyle\sum_{i=1}^{n}f\left (x_i^*\right)\Delta x$
$\Delta x=\frac{(b-a)}{n}$
$x_i^*=a+(\Delta x)i$

Then:
$\Delta x=\frac{\left(1-0\right)}{n}=\frac{1}{n}$
$\frac{1+2+3+...+n}{n^2+n}=\displaystyle\sum_{i=1}^ {n}\frac{i}{n^2+n}=\displaystyle\sum_{i=1}^{n}\fra c{i}{n+1}\cdot\frac{1}{n}$

When I decompose the sum into $\displaystyle\sum_{i=1}^{n}\frac{i}{n+1}\cdot\frac {1}{n}$, I think I'm seing a basis of a Riemann sum with the following:

$f\left(x_i^*\right)=\frac{i}{n+1}$
$\Delta x=\frac{1}{n}$

If I'm correct, then I continue...

$\displaystyle\sum_{i=1}^{n}\frac{i}{n+1}\cdot\frac {1}{n}=\frac{1}{n}\left(\frac{1}{n+1}\cdot\display style\sum_{i=1}^{n}i\right)=\frac{1}{n}\left(\frac {1}{n+1}\cdot\frac{n(n+1)}{2}\right)=\frac{1}{n}\c dot\frac{n}{2}=\frac{1}{2}$

Assuming all of that is correct, the lim as n approaches infinity is 1/2. However, I've spent a lot of time looking at this problem and might be completely off track.

Thanks,
Manny
• August 11th 2009, 10:44 AM
Failure
Quote:

Originally Posted by manny
Evaluate $\lim_{n\to\infty}\frac{1+2+3+...+n}{n^2+n}$ by first recognizing the sum as a Riemann sum over a partition of [0,1] and then evaluating the corresponding integral.
A. 0
B. 1/2
C. 1/4
D. Not enough information
E. None of the above
----------------------------------------------------------------------------------------------------------------

What is confusing me about the question is whether I'm being asked to compute a Riemann sum or being asked to "reverse engineer" the given function into a Riemann sum. I tend to lean towards the latter so this is what I've done so far...

Given:
$\displaystyle\int^b_a f\left(x\right)\,dx = \lim_{n\to\infty}\displaystyle\sum_{i=1}^{n}f\left (x_i^*\right)\Delta x$
$\Delta x=\frac{(b-a)}{n}$
$x_i^*=a+(\Delta x)i$

Then:
$\Delta x=\frac{\left(1-0\right)}{n}=\frac{1}{n}$
$\frac{1+2+3+...+n}{n^2+n}=\displaystyle\sum_{i=1}^ {n}\frac{i}{n^2+n}=\displaystyle\sum_{i=1}^{n}\fra c{i}{n+1}\cdot\frac{1}{n}$

When I decompose the sum into $\displaystyle\sum_{i=1}^{n}\frac{i}{n+1}\cdot\frac {1}{n}$, I think I'm seing a basis of a Riemann sum with the following:

$f\left(x_i^*\right)=\frac{i}{n+1}$
$\Delta x=\frac{1}{n}$

If I'm correct, then I continue...

Great, so far. But now you go off in the wrong direction insofar as you are asked, not to directly evaluate the limit of these Riemann sums as $n\rightarrow \infty$, but just to evaluate the corresponding integral.

$\int_0^2 x\, dx=\tfrac{1}{2}x^2\big|_{x=0}^1=\tfrac{1}{2}$

Quote:

$\displaystyle\sum_{i=1}^{n}\frac{i}{n+1}\cdot\frac {1}{n}=\frac{1}{n}\left(\frac{1}{n+1}\cdot\display style\sum_{i=1}^{n}i\right)=\frac{1}{n}\left(\frac {1}{n+1}\cdot\frac{n(n+1)}{2}\right)=\frac{1}{n}\c dot\frac{n}{2}=\frac{1}{2}$

Assuming all of that is correct, the lim as n approaches infinity is 1/2. However, I've spent a lot of time looking at this problem and might be completely off track.

Thanks
The result seems ok (to me, that is). But you have done more work than you were asked to do. In other cases, it may be a terrible amount of work to evaluate the limit of the Riemann sums directly, instead of evaluating the corresponding integral.
• August 11th 2009, 10:51 AM
red_dog
$a_n=\frac{1}{n}\sum_{k=1}^n\frac{k}{n+1}=\frac{n+1 }{n}\cdot\frac{1}{n+1}\sum_{k=0}^n\frac{k}{n+1}$

The Riemann sum is $\frac{1}{n+1}\sum_{k=0}^n\frac{k}{n+1}$.

In this case $\Delta x=\frac{1}{n+1}, \ x_k^*=\frac{k}{n+1}, \ k=0,1,\ldots,n$ and the function is $f:[0,1]\to\mathbb{R}, \ f(x)=x$

Then, $\lim_{n\to\infty}a_n=\lim_{n\to\infty}\frac{n+1}{n }\cdot\lim_{n\to\infty}\frac{1}{n+1}\sum_{k=0}^n\f rac{k}{n+1}=1\cdot\int_0^1xdx=\left.\frac{x^2}{2}\ right|_0^1=\frac{1}{2}$
• August 11th 2009, 11:20 AM
manny
Quote:

Originally Posted by Failure
Great, so far. But now you go off in the wrong direction insofar as you are asked, not to directly evaluate the limit of these Riemann sums as $n\rightarrow \infty$, but just to evaluate the corresponding integral.

$\int_0^2 x\, dx=\tfrac{1}{2}x^2\big|_{x=0}^1=\tfrac{1}{2}$

The result seems ok (to me, that is). But you have done more work than you were asked to do. In other cases, it may be a terrible amount of work to evaluate the limit of the Riemann sums directly, instead of evaluating the corresponding integral.

It seems you're right and I have done extra work. However, I can't see how you got $\int_0^1 x\, dx$, which forces me to do it the long way. Would you mind clarifying, basically, how you saw/know that f(x)=x? I think once I understand that, the entire problem will make sense because I really struggled in the beginning to determine what the actual function was in the question.

Thanks to both Failure and red_dog for the help/patience thus far,
Manny
• August 13th 2009, 03:35 PM
manny
Bump. Anyone able to explain how to get $\int_0^1 x\, dx$ like red_dog and Failure did?

-Manny
• August 13th 2009, 11:01 PM
red_dog
In the Riemann sum replace $x_k^*$ with $x$ to find $f(x)$

In this case $f(x_k^*)=\frac{k}{n+1}=x_k^*$.

Replacing $x_k^*$ with $x$ we get $f(x)=x$
• August 14th 2009, 07:27 PM
manny
Thanks red_dog. I ended up stumbling into the answer a while later but appreciate the follow up.

Manny