# Infinite series (i think it's riemann)

• flyerpower
In summary: Then I have made a mistake cause it get to be 0.373551 and not a 1/3 :(I may have been wrong but wolframalpha says it's 1/3 so it must be correct :).Use those bounds and take the limits, you'll see they both converge to 1/3.
flyerpower

## Homework Statement

$\sum_{1}^{inf}$ k^2/(n^3+k^2)

## The Attempt at a Solution

I think it's Riemann but i cannot find a suitable function to integrate.

flyerpower said:

## Homework Statement

$\sum_{1}^{inf}$ k^2/(n^3+k^2)

## The Attempt at a Solution

I think it's Riemann but i cannot find a suitable function to integrate.

Isn't it suppose to be

$$\sum_{n = 1}^{\infty} \frac{k^2}{n^3 + k^2}$$

Please take more care to expressing to others (and yourself) what it is you want to find. As written the sum doesn't make total sense. It could be what Susanne217 said above, or Riemann sum comment makes me think you could have also meant $\displaystyle\lim_{n\to\infty} \sum_{k=1}^n \frac{k^2}{n^3 + k^2}$.

Gib Z said:
Please take more care to expressing to others (and yourself) what it is you want to find. As written the sum doesn't make total sense. It could be what Susanne217 said above, or Riemann sum comment makes me think you could have also meant $\displaystyle\lim_{n\to\infty} \sum_{k=1}^n \frac{k^2}{n^3 + k^2}$.

I agree...

I have been away for a while but I think it could be that the young man is suppose re-write the sum to an improper integral and thereby finding the sum of the series??

In the case that it is what I thought, then it's not as simple as recognizing it as a pre-prepared Riemann sum. With some careful estimates to bound the sum, you should get the result to be $1/3$.

Gib Z said:
In the case that it is what I thought, then it's not as simple as recognizing it as a pre-prepared Riemann sum. With some careful estimates to bound the sum, you should get the result to be $1/3$.

Isn't it true that you need to rewrite this series into more easy to work with??

First of all sorry for misspelling.

I used the bounds k^2/(n^3+n^2) <= k^2/(n^3+k^2) <= k^2/(n^3+1) and i worked it out to 1/3.

flyerpower said:
First of all sorry for misspelling.

I used the bounds k^2/(n^3+n^2) <= k^2/(n^3+k^2) <= k^2/(n^3+1) and i worked it out to 1/3.

Then I have made a mistake cause it get to be 0.373551 and not a 1/3 :(

I may have been wrong but wolframalpha says it's 1/3 so it must be correct :).
Use those bounds and take the limits, you'll see they both converge to 1/3.

## 1. What is an infinite series?

An infinite series is a sum of infinitely many terms. Each term is added to the previous one, creating a potentially infinite sequence of numbers.

## 2. How does a Riemann sum differ from an infinite series?

A Riemann sum is a finite approximation of an infinite series. It uses a specific number of terms to estimate the value of the infinite series.

## 3. What is the Riemann zeta function?

The Riemann zeta function is a mathematical function that represents the sum of the reciprocals of all positive integers raised to a given power.

## 4. How is the Riemann hypothesis related to infinite series?

The Riemann hypothesis is a conjecture in mathematics that suggests a connection between the distribution of prime numbers and the behavior of infinite series, specifically the Riemann zeta function.

## 5. What are some real-world applications of infinite series?

Infinite series have many applications in physics, engineering, and other fields. They can be used to model natural phenomena, solve differential equations, and optimize systems.

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