Riemann Sums question

[ksdhart]

Hello all. I am having difficulty with a Riemann Sum problem from my Calculus I course. The problem asks me to the compute the value of this Riemann sum:

\(\displaystyle \displaystyle \lim_{n\to\infty} \sum_{k=1}^n {k^3 \over n^4+n+1}\)

The way I went about solving this problem was to break the sum into three separate sums and then pull out the constants

\(\displaystyle \displaystyle \lim_{n\to\infty} {1 \over n^4} \times \sum_{k=1}^n {k^3} + {1 \over n} \times \sum_{k=1}^n {k^3} + {1 \over 1} \times \sum_{k=1}^n {k^3}\)

I have a handy sum formula for k³ ... \(\displaystyle \displaystyle \sum_{k=1}^n {k^3} = {n^2 \times (n+1)^2 \over 4} = {n^4 + 2n^3 + n^2 \over 4}\) ... and so I can turn the problem into

\(\displaystyle \displaystyle \lim_{n\to\infty} {n^4 + 2n^3 + n^2 \over 4n^4} + {n^4 + 2n^3 + n^2 \over 4n} + {n^4 + 2n^3 + n^2 \over 4}\)

Further breaking apart into separate fractions then leaves me with

\(\displaystyle \displaystyle \lim_{n\to\infty} {n^4 \over 4n^4} + {2n^3 \over 4n^4} + {n^2 \over 4n^4} + {n^4 \over 4n} + {2n^3 \over 4n} + {n^2 \over 4n} + {n^4 \over 4} + {2n^3 \over 4} + {n^2 \over 4}\)

Which is simplified to

\(\displaystyle \displaystyle \lim_{n\to\infty} {1 \over 4} + {1 \over 2n} + {1 \over 4n^2} + {n^3 \over 4} + {n^2 \over 2} + {n \over 4} + {n^4 \over 4} + {2n^3 \over 4} + {n^2 \over 4}\)

At this point, I have a major problem. I have n terms in the numerator and so the sum goes to infinity

\(\displaystyle 0.25 + 0 + 0 + \infty + \infty + \infty + \infty + \infty + \infty = \infty\)

I know I went wrong somewhere along the line, that much is obvious. But I cannot for the life of me figure out where. Can anybody help me out, please?

 

[Ishuda]

Hello all. I am having difficulty with a Riemann Sum problem from my Calculus I course. The problem asks me to the compute the value of this Riemann sum:

\(\displaystyle \displaystyle \lim_{n\to\infty} \sum_{k=1}^n {k^3 \over n^4+n+1}\)

The way I went about solving this problem was to break the sum into three separate sums and then pull out the constants

\(\displaystyle \displaystyle \lim_{n\to\infty} {1 \over n^4} \times \sum_{k=1}^n {k^3} + {1 \over n} \times \sum_{k=1}^n {k^3} + {1 \over 1} \times \sum_{k=1}^n {k^3}\)
...

This part is incorrect
\(\displaystyle \displaystyle \lim_{n\to\infty} {1 \over n^4} \times \sum_{k=1}^n {k^3} + {1 \over n} \times \sum_{k=1}^n {k^3} + {1 \over 1} \times \sum_{k=1}^n {k^3}\)
That is
\(\displaystyle (\frac{1}{n^4}+\frac{1}{n}+\, 1)\, \Sigma\, k^3\,\, =\, \frac{1\, +\, n^3\, +\, n^4}{n^4} \Sigma\, k^3\)
You could just factor out the \(\displaystyle \frac{1}{n^4\, +\, n\, +\, 1}\) from the original equation.

 

[ksdhart]

Thank you very very much. I'm embarrassed to admit that I apparently forgot how fractions work. I swear that since taking college-level math courses, my brain has atrophied and I've forgotten everything I ever learned in basic math. At any rate, once I did the problem correctly, I ended up with a limit that goes to an indeterminate form, so I used L'Hopital's Rule a few times until it simplified down to the correct answer of 1/4.

 

[Ishuda]

Thank you very very much. I'm embarrassed to admit that I apparently forgot how fractions work. I swear that since taking college-level math courses, my brain has atrophied and I've forgotten everything I ever learned in basic math. At any rate, once I did the problem correctly, I ended up with a limit that goes to an indeterminate form, so I used L'Hopital's Rule a few times until it simplified down to the correct answer of 1/4.

Another way to approach the limit other than L'Hopital's Rule is the following
\(\displaystyle \dfrac{1}{n^4 + n + 1} \underset{1}{\overset{n}{\Sigma}} \, k^3\)
\(\displaystyle =\dfrac{1}{n^4 + n + 1} \dfrac{n^4 + 2 n^3 + n^2}{4}\)
\(\displaystyle =\dfrac{n^4 + 2 n^3 + n^2}{4 (n^4 + n + 1)}\)
\(\displaystyle =\dfrac{n^4 ( 1 + \dfrac{2}{n} + \dfrac{1}{n^2})}{4 n^4 (1 + \dfrac{1}{n^3} + \dfrac{1}{n^4})}\)
\(\displaystyle =\dfrac{1}{4}\, \dfrac{1 + \dfrac{2}{n} + \dfrac{1}{n^2}}{1 + \dfrac{1}{n^3} + \dfrac{1}{n^4}}\)
The limit (as n goes to infinity) of that last term is one so the complete limit is 1/4.

Oh, and as I have said before, "If you feel like knocking yourself up side the head, don't! It hurts, as I know from personal experience."