Question

So I am to use an "appropriate Riemann sum" to evaluate this: \[\lim_{n \to \infty} \frac{1^5 + 2^5 + 3^5 . . .n^5}{n^6}\]
If anyone could help out I'd appreciate it. And, if possible, give me an idea of how to generally write a Riemann sum from an expression like this, because I always have trouble doing it. Here's my attempt for this:

Answer 1

\[\lim_{n \to \infty} \frac{1^5 + 2^5 + 3^5 . . .n^5}{n^6} = \lim_{n \to \infty} \frac{i^5 +n^5}{n^5} \cdot \frac{1}{n}\]
\[\frac{1}{n}\sum_{i=1}^n\frac{i^5 + n^5}{n^5} = \frac{1}{n} \cdot \frac{1}{n^5} \sum_{i=1}^n [i^5 + 1]\]
\[\frac{1}{n^6} \cdot \left(\frac{n^2(n+1)^2(2n^2+2n-1)}{12} +n\right)\]
\[\lim_{n \to \infty} \frac{2n^6 +6n^5+5n^4-x^2+12n}{12n^6} = \frac{1}{6}\]

Answer 2

how about this :

\[\begin{align}\lim_{n \to \infty} \frac{1^5 + 2^5 + 3^5 . . .n^5}{n^6} &= \lim_{n \to \infty} \frac{1}{n^6}\sum\limits_{i=1}^n i^5 \\~\\
&= \lim_{n \to \infty} \sum\limits_{i=1}^n \left(\frac{i}{n}\right)^5\cdot \frac{1}{n} \\~\\
&= \lim_{n \to \infty} \sum\limits_{i=1}^n \left(\frac{i}{n}\right)^5\cdot \frac{1-0}{n} \\~\\
&= \int\limits_0^1 x^5 \, dx\\~\\
&= \frac{1}{6}

\end{align}\]

Answer 3

Well it looks like we both got the same answers. I understand how to do definite integrals, but the question asks to use a Riemann sum to evaluate it. But looking at your work, I see how to find the domain of the function, which I didn't know how to do before :)

Answer 4

slick

Answer 5

abb1"t, I don't understand

Answer 6

what do you know about the relationship between riemann sum and definite integral ?

Answer 7

I know that the definite integral is the (exact) area under a function in a certain domain, and that the riemann sum achieves the same thing by finding the limit as n approaches infinity of the sum of the function on the domain, where n is the number of rectangles. I've done some exercises involving this:

\[\lim_{n \to \infty} \sum_{i=1}^n f(c_i) \Delta x = \int_a^b \! f(x)\ dx\]

Answer 8

Right, `use an "appropriate Riemann sum" to evaluate this: `

doesn't that mean to write the given expression as riemann sum so the the limit of that equals the defintie integral which you can evaluate easily ?

Answer 9

\[\begin{align}\lim_{n \to \infty} \frac{1^5 + 2^5 + 3^5 . . .n^5}{n^6} &= \lim_{n \to \infty} \frac{1}{n^6}\sum\limits_{i=1}^n i^5 \\~\\
&= \lim_{n \to \infty} \sum\limits_{i=1}^n \left(\frac{i}{n}\right)^5\cdot \frac{1}{n} \\~\\
&= \lim_{n \to \infty} \color{Red}{\sum\limits_{i=1}^n \left(\frac{i}{n}\right)^5\cdot \frac{1-0}{n} }\\~\\
&= \int\limits_0^1 x^5 \, dx\\~\\
&= \frac{1}{6}

\end{align}\]

that red part is the riemann sum that the question is asking you to use

Answer 10

here you can find couple of similar problems and solutions
http://www.math.toronto.edu/canghel/mat135Q/Q5_3.pdf

Answer 11

That makes sense; I just interpreted it as them wanting me to use the Riemann sum all the way through instead of finding the sum and then evaluating it using the definite integral, because they had me finding definite integrals by finding the limit of the sum before they actually taught me definite integrals. So, would it be INCORRECT to solve the quesion as I did? or is it just a "finding the limit of the sum is way harder" type of thing?

Answer 12

there is nothing wrong in you approach in general except for few mistakes

but you're not using riemann sum -
riemann sum refers to approximation of area under curve. I feel, the question as it stands, wants you to use riemann sum and evaluate the definite integral...

Answer 13

I see you're cleverly using sum of fifth powers of first n natural numbers formula to evaluate the limit and i like it more than the riemann sum thingy xD

Answer 14

Haha, thanks but I can't take the credit for it; it was one of the methods in my textbook. Do you mind pointing out the mistakes I made in my original work?

Answer 15

but there are few obvious mistakes in your work which im sure you will see if you go through it again

Answer 16

lets look at first line of your work : \[\lim_{n \to \infty} \frac{1^5 + 2^5 + 3^5 . . .n^5}{n^6} = \lim_{n \to \infty} \frac{i^5 +n^5}{n^5} \cdot \frac{1}{n}\]

Answer 17

does this jump really make sense ?

Answer 18

what is \(i\) here ?

Answer 19

I left out some details, I meant for \(i\) to be the variable in the sum since that's what is changing in the pattern. I was trying to manipulate it to fit the definition of a riemann sum:

\[\sum_{i = 1}^n f(c_i) \cdot \Delta x\]

Answer 20

adding that detail wont make that first line true

Answer 21

lets pause on that and look at second line :
\[\frac{1}{n}\sum_{i=1}^n\frac{i^5 + \color{Red}{n^5}}{n^5} = \frac{1}{n} \cdot \frac{1}{n^5} \sum_{i=1}^n [i^5 + \color{red}{1}]\]


can you explain how that n^5 became 1 ?

Answer 22

im just pointing the mistakes i spotted to you because you have asked me to :D

Answer 23

I tried to expand the fraction to be \(\frac{i^5}{n^5} + \frac{n^5}{n^5}.\)

\[\frac{1}{n} \sum_{i=1}^n \frac{i^5}{n^5} + \frac{n^5}{n^5}\]
\[= \frac{1}{n} \sum_{i=1}^n \frac{i^5}{n^5} + 1 \]
\[= \frac{1}{n} \cdot \frac{1}{n^5} \sum_{i=1}^n i^5 + 1\]

Answer 24

And it's not problem, I know I asked you XD I won't get offended, promise

Answer 25

ikr :) so that "1" no longer is attached to SUM ? beacuse i see you're not using parenthesis anymore

Answer 26

Sorry, got lazy. It is still attached to the sum. It should be

\[\sum_{i=1}^n[i^5 + 1]\]

Answer 27

if you put parenthesis, you cannot factor out 1/n^5

Answer 28

\[\frac{1}{n}\sum_{i=1}^n\frac{i^5 + n^5}{n^5} \ne \frac{1}{n} \cdot \frac{1}{n^5} \sum_{i=1}^n [i^5 + 1]\]

Answer 29

Any particular reason why? I was taught that you could factor out constants from a sum. And we treat the \(n\) as a constant because it represents the end of the pattern, i.e. it never changes.

Answer 30

notice you're just pulling out n^5 from the denominator, so numerator has nothing to do with it. it stays the same :

\[\frac{1}{n}\sum_{i=1}^n\frac{i^5 + n^5}{n^5} = \frac{1}{n} \cdot \frac{1}{n^5} \sum_{i=1}^n [i^5 + n^5]\]

Answer 31

what im saying is there is no way for that n^5 to become 1 when you pull out n^5

Answer 32

Also keep in mind we're messing with a wrong expression because there is a mistake carried on from first line

Answer 33

Okay, so what you're saying is that I accidently got the right answer. XD

Answer 34

:P what im saying is today is your lucky day, go do all your important things haha !

Answer 35

Here I have fixed the mistakes that we discussed and updated your work :
\[\lim_{n \to \infty} \frac{1^5 + 2^5 + 3^5 . . .n^5}{n^6} = \lim_{n \to \infty} \sum\limits_{i=1}^n\frac{i^5}{n^6} \]

\[= \lim_{n \to \infty} \frac{1}{n^6} \sum_{i=1}^n i^5 \]

\[= \lim_{n \to \infty} \frac{1}{n^6} \cdot \left(\frac{n^2(n+1)^2(2n^2+2n-1)}{12}\right)\]

\[= \lim_{n \to \infty} \frac{2n^6 +6n^5+5n^4-n^2}{12n^6} = \frac{1}{6}\]

see if it makes sense more or less :)

Answer 36

Yup, I believe I understand it except for one question. You express the original expression as

\[\lim_{n \to \infty} \sum_{i=1}^n \frac{i^5}{n^6}.\]
Is the numerator \(i^5\) because the variable \(i\) accounts for EVERY change in the series, including \(n\)?

Answer 37

Because I thought \(n\) was a separate entity

Answer 38

1^5+2^5......+n^5=sum[i=1 to n] i^5
the denominator has nothing to do with that summations
the sum [i=1 to n] (i^5) is summing all positive integers raised to 5 all the way to n

Answer 39

i must you did a good to job getting the answer lol

Answer 40

say

Answer 41

did a good job*

Answer 42

Thanks for your help xapproachesinfinity :) I think I understand it now

Answer 43

welcome!