Infinite Series Test: 1/n^2 - 1/n^3

In summary, the given series is a telescoping series and when evaluated, the terms do not cancel out. Therefore, the series is divergent. It would be helpful to know what the intended purpose of the series is and what tests can be used to determine its convergence or divergence.
  • #1
renob
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Homework Statement


[tex]\sum[/tex][tex]^{infinity}_{n=1}[/tex] (1/n^2 - 1/n^3)

Homework Equations


it goes to infinity
n=1


The Attempt at a Solution


Im assuming this is a telescoping series. when I plugged in my terms nothing canceled out except for the 1's at the beginning.

(1-1)+(1/4 - 1/8)+(1/9 - 1/27)+...

Am I doing something wrong? Or does it just go to infinity and therefore diverge?
 
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  • #2
You didn't say what you're supposed to do with this series, which would have been useful information.

Presumably you're supposed to determine that the series is divergent or it is convergent. What tests do you know of that you can use?

When you ask "does it just go to infinity", what are you referring to by "it"? Whatever "it" is, why do you think "it" goes to infinity? There's nothing in the work you show that suggests that to me.
 

What is an infinite series test?

An infinite series test is a mathematical technique used to determine whether an infinite series (a sum of infinitely many terms) converges or diverges.

What is the formula for the infinite series 1/n^2 - 1/n^3?

The formula for the infinite series 1/n^2 - 1/n^3 is 1/n^2 - 1/n^3 = 1/n^2(1 - 1/n).

How do you determine if the infinite series 1/n^2 - 1/n^3 converges or diverges?

To determine if the infinite series 1/n^2 - 1/n^3 converges or diverges, we can use the limit comparison test. If we take the limit as n approaches infinity of the ratio of the given series to the corresponding p-series (1/n^2), and the limit is a non-zero finite number, then the given series converges. Otherwise, it diverges.

What is the significance of the value of the series for n=1?

The value of the series for n=1 is 0. This means that the first term of the series (1/1^2 - 1/1^3 = 0) does not contribute to the overall sum of the series.

Can the infinite series 1/n^2 - 1/n^3 be used to approximate other series?

Yes, the infinite series 1/n^2 - 1/n^3 can be used to approximate other series, such as the alternating harmonic series. By rearranging the terms of the given series, we can create a new series that has the same sum but converges more quickly.

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