Determine whether the series is convergent or divergent

In summary, the series is convergent, but not sure how to prove it mathematically. The attempt at a solution suggests that the series might be expressed as a sum of a p-series, and that the numerator and denominator are both convergent.
  • #1
tnutty
326
1

Homework Statement


Determine whether the series is convergent or divergent.

1 + 1/8 + 1/27 + 1/64 + 1/125 ...


Homework Equations





The Attempt at a Solution



I know this is convergent but not sure how to prove this mathematically.
 
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  • #2
It might help to write the series in it's summation form...

Do you know any of the tests for convergence?
 
  • #3
Well this chapter is about the integral test so can I apply that method?
 
  • #4
Yeah you can apply that test, I suppose -- How does the integral test work?
 
  • #5
i don't think the integral test can apply to this because this cannot be represented as a function.

however, there is another test, the remainder test,

Remainder test states that :

Suppose f(k) = ak, where f is a continuous,positive,decreasing function for x>=n
and [tex]\sum a_n [/tex] is convergent. If Rn= s-sn,

then [tex]\int f(x) dx[/tex] [tex]\leq R_n[/tex] [tex]\leq \int f(x) dx[/tex], where the limit is from n to infinity
 
  • #6
I think you're misunderstanding the integral test

It says that [tex]\sum_{n=1}^{infinity} f(n)[/tex] converges if [tex]\int_{n}^{infinity} f(x) dx[/tex] converges.
 
  • #7
tnutty said:

Homework Statement


Determine whether the series is convergent or divergent.

1 + 1/8 + 1/27 + 1/64 + 1/125 ...

Homework Equations


The Attempt at a Solution



I know this is convergent but not sure how to prove this mathematically.
Through some induction it can be seen that the summation is: [tex]$\displaystyle\sum_{n=1}^\infty \frac{1}{n^3}.[/tex]

Which is in the form of a p-series:

[tex]$\displaystyle\sum_{n=1}^\infty \frac{1}{n^p}.[/tex]

So if you were to know the rules of the p-series I assume it would be pretty straightforward
 
  • #8
Ok, I get that, but how can we represent the problem above as a sum of a series, so
we can use the integral test?
 
  • #9
Do you know what the general term (i.e., nth term) in your series looks like?

1 + 1/8 + 1/27 + 1/64 + 1/125 ... + ? + ...

There's a definite pattern going on here.
 
  • #10
Ya that's the problem, i can't seem to recognize any patterns.
 
  • #11
EDIT: In helping you recognize the pattern, you should first look at the numerator...

The numerator is always 1, so you can sort of look beyond that and concentrate on the denominator...

Since the first value of the series is 1, you can surmise that the summation probably begins at [tex]n=1[/tex] or has a number raised to the [tex]n[/tex] value (which if n began at 0 would yield 1). In this case however, beginning at 0 doesn't make sense as the following terms don't correspond with a series beginning at an index of 0.

Now that you can concentrate on the denominator, you can't do much more than use induction to realize what each of the denominators has in common. In this case, each of the 5 terms' denominators, (1, 8, 27, 64, 125) is a perfect cube number. Once you "see" this, try to plug in your index and the succeeding values of the series to see if they would match up with [tex]n^3[/tex].

In this case they do! Thus this series can be expressed as simply: [tex]$\displaystyle\sum_{n=1}^\infty \frac{1}{n^3}.[/tex]

I hope this helps!

tnutty said:
Ok, I get that, but how can we represent the problem above as a sum of a series, so
we can use the integral test?

If you insist on using the Integral test on this just take the integral of the summation we've previously noted:
[tex]$\displaystyle\sum_{n=1}^\infty \frac{1}{n^3}.[/tex]

For example:

[tex]\displaystyle\int^\infty_1 n^{-3}dn[/tex]

This maintains the convergence you've already established.
 
Last edited:
  • #12
Oh man good eye, I just got off spring break and have been trying hard to get back into
the mode of thinking. Thanks man!.
 

1) What is the difference between a convergent and divergent series?

A convergent series is one in which the sum of its terms approaches a finite limit as the number of terms increases. In contrast, a divergent series is one in which the sum of its terms either does not approach a limit or approaches a limit of infinity.

2) How do you determine whether a series is convergent or divergent?

One method is to use the ratio test, where the limit of the absolute value of the ratio of consecutive terms is taken. If this limit is less than 1, the series is convergent. If the limit is greater than 1, the series is divergent. If the limit is equal to 1, the test is inconclusive and other methods, such as the comparison test, may need to be used.

3) What is the purpose of determining whether a series is convergent or divergent?

Determining the convergence or divergence of a series is important in many areas of mathematics, such as calculus and differential equations. It allows us to understand the behavior of the series and make accurate predictions and calculations based on its properties.

4) Are there any other tests besides the ratio test for determining convergence or divergence?

Yes, there are several other tests that can be used, such as the comparison test, the integral test, and the alternating series test. Each test has its own set of conditions and is useful in different scenarios.

5) Can a series be both convergent and divergent?

No, a series can only be either convergent or divergent, not both. However, there are some series that are conditionally convergent, meaning that they converge under certain conditions but diverge under others. These types of series can be more complicated to analyze and may require additional tests to determine their convergence or divergence.

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