- #1

- 22

- 0

If the serie

∞

∑ Xn = (1/n²)

n=1

converge, then the serie |Xn| converge?

Second question

the serie

∞

∑ X^n / [fat(n)] diverge?

n=1

where fat(n) = n(n-1)(n-2)...4.3.2.1

Its important know the value of X in X^n?

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- Thread starter oswald
- Start date

In summary, the first conversation is discussing the convergence of a series (1/n²) and its absolute value, both of which converge. The second conversation is discussing the convergence of a series (X^n / [fat(n)]) and it is suggested to use a ratio test which will show that the value of X does not affect the convergence.

- #1

- 22

- 0

If the serie

∞

∑ Xn = (1/n²)

n=1

converge, then the serie |Xn| converge?

Second question

the serie

∞

∑ X^n / [fat(n)] diverge?

n=1

where fat(n) = n(n-1)(n-2)...4.3.2.1

Its important know the value of X in X^n?

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- #2

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https://www.physicsforums.com/showthread.php?t=94383

- #3

- 173

- 0

oswald said:

If the serie

∞

∑ Xn = (1/n²)

n=1

converge, then the serie |Xn| converge?

Second question

the serie

∞

∑ X^n / [fat(n)] diverge?

n=1

where fat(n) = n(n-1)(n-2)...4.3.2.1

Its important know the value of X in X^n?

1)what is the || of 1/n^2?? think

2) search for geometric series and the hierarchy of infinities..

regards

marco

- #4

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is a well known Taylor's series for a simple function and your sum is only slightly different.

- #5

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1)

∞

∑ Xn = (1/n²) = 1 + 1/4 + 1/9 ... its a p-serie and converge because p=2>1.

n=1

and

∞

∑ |Xn| = (1/n²) = 1 + 1/4 + ... its the same.. so its converge too.

n=1

2) converge using comparison test, no dought!

- #6

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oswald said:

1)

∞

∑ Xn = (1/n²) = 1 + 1/4 + 1/9 ... its a p-serie and converge because p=2>1.

n=1

and

∞

∑ |Xn| = (1/n²) = 1 + 1/4 + ... its the same.. so its converge too.

n=1

2) converge using comparison test, no dought!

Better to use a ratio test. It will make it easy to see why the value of x doesn't matter.

The concept of series convergence/divergence is used in mathematics to determine whether a given infinite series (a sum of an infinite number of terms) has a finite value or not. If a series is convergent, it means that the sum of its terms approaches a finite value as the number of terms increases. On the other hand, if a series is divergent, it means that the sum of its terms approaches infinity as the number of terms increases.

There are several tests that can be used to determine whether a series is convergent or divergent. These include the ratio test, the comparison test, the integral test, and the alternating series test. Each of these tests has specific criteria and conditions that must be met for the series to be classified as convergent or divergent.

The concept of series convergence/divergence has practical applications in various areas of science and engineering. For example, it is used in physics to calculate the behavior of electric and magnetic fields, in chemistry to determine the rate of a chemical reaction, and in economics to analyze financial data and predict future trends.

No, a series cannot be both convergent and divergent. A series can only have one of these two classifications, and it cannot switch between them. If a series satisfies the criteria for both convergence and divergence, it is considered to be indeterminate, which means that more advanced mathematical techniques are needed to determine its behavior.

Series convergence/divergence plays a crucial role in many areas of mathematics and science. It allows us to analyze and understand the behavior of infinite sequences and series, which are essential in many mathematical and scientific models. Additionally, understanding series convergence/divergence can help us make predictions and draw conclusions about real-world phenomena.

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