Alternating series something dissapeared.

In summary, the alternating series $$A_n = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n+1}$$ converges by the Leibniz criterion because the sequence it is formed from, a_n=1/(n+1), is monotonically decreasing and goes to 0 as n approaches infinity. By using the trick of generating functions and integrating from 0 to 1, the value of the series can be found to be 1-ln(2). However, this method only works if the series actually converges. Otherwise, it falls under the category of regularization and would need to be proven separately.
  • #1
Jbreezy
582
0

Homework Statement



Hi,
The question wanted to know if the alternating series converges or diverges.

$$A_n = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n+1}$$

Homework Equations





3. The Attempt at a Solution [/b
You can see it here .

http://www.calcchat.com/book/Calculus-ETF-5e/ chapter 9, section 5, question 11
So I understand the process I just want to know why they ignore the $$ (-1)^{n+1}$$ in the numerator all of a sudden it just becomes 1.


I understand that it will be either 1 or -1 depending on the value of n. But why can they just ignore if it was -1 and say it is 1 and take the limit? I don't get it.
 
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  • #2
The statement of theorem 9.14 is going to start with: suppose you have a sum of the form
[tex] \sum (-1)^{n+1} a_n [/tex]

Where the an are positive and decreasing.

So the theorem says if you have an alternating series, you can take the absolute value of the terms to show that it converges (usually)
 
  • #3
Theorem 9.14

Let $$A_n >0 $$.
The alternating series

$$\sum (-1)^{n} a_n$$ and $$\sum (-1)^{n+1} a_n$$
converge if the following two conditions are met.

1. lim --> ∞ $$A_n = 0$$ 2. $$A_n+1 <= A_n for all n$$
 
  • #4
Ahh crap for 2 I meant A_(n+1) <= A_n for all n

I suck at the writing the equations. But yeah no abs anywhere in that theorm
 
  • #5
All the theorem is trying to say is that if the [itex]a_n[/itex] sequence is decreasing, then the alternating series is convergent.

In this case, the series is convergent, and more specifically, it converges to [itex]1-\log(2)[/itex].
 
  • #6
Jbreezy said:

Homework Statement



Hi,
The question wanted to know if the alternating series converges or diverges.

$$A_n = \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n+1}$$

Homework Equations


3. The Attempt at a Solution [/b
You can see it here .

http://www.calcchat.com/book/Calculus-ETF-5e/ chapter 9, section 5, question 11
So I understand the process I just want to know why they ignore the $$ (-1)^{n+1}$$ in the numerator all of a sudden it just becomes 1.I understand that it will be either 1 or -1 depending on the value of n. But why can they just ignore if it was -1 and say it is 1 and take the limit? I don't get it.


Theorem 9.14 is a statement of a test for convergence called the "alternating series test", also known as the Leibniz criterion. It applies ONLY to alternating series, which have terms that oscillate between positive and negative from one term to the next. It does NOT apply to series where all the terms are positive or all the terms are negative, or even when the positive and negative terms don't "neatly" alternate from one term to the next.

This criterion requires that (for such an alternating series to converge) when you disregard the sign and just take the absolute value (only the numerical magnitude) of each term, these values go monotonically to zero at the limit.

Read that carefully. You need to prove that there's a steady trend where the next value is lower numerically than the one before, and that the "last term" (limit) goes to zero. Theorem 9.14 puts that in mathematical notation.

When you list the absolute values of the terms in this series, they in fact form part of the harmonic sequence omitting 1, viz.: ##\displaystyle \frac{1}{2}, \frac{1}{3}, \frac{1}{4},...,\frac{1}{k},...##. The proof is just showing that it's always true that each term in this sequence is smaller than the one before; and further that the limit of the sequence goes to zero. Both of these are trivial, and you should easily be able to see these.

This should be sufficient to prove that the alternating series formed from these terms converges (based on the Leibniz criterion). However, even though we brought in the absolute values of the terms, don't get confused into calling the series "absolutely convergent", because it is NOT absolutely convergent. If you take all those absolute values and add them together, you get the harmonic series minus one, i.e. ##\displaystyle H - 1##, which is clearly divergent. It is purely because this is an alternating series formed with those same terms that you can use a "weaker" criterion (like the Leibniz criterion) to show convergence.
 
Last edited:
  • #7
It's convergent, because the squence [itex]a_n=1/(n+1)[/itex] is monotonically decreasing and going to 0 for [itex]n \rightarrow \infty[/itex]. Then according to the Leibniz criterion the alternating series converges.

To get the value, use the trick of generating functions and define
[tex]f(x)=\sum_{n=1}^{\infty} (-x)^n.[/tex]
This series is convergent for [itex]|x|<1[/itex]. Since it's a geometric series you can sum it explicitly
[tex]f(x)=\frac{1}{x+1}-1.[/tex]
Now you get your series by integrating [itex]f(x)[/itex] from 0 to 1 (argue why!):
[tex]\int_0^1 \mathrm{d} x f(x)=[\ln(x+1)-x]_{x=0}^{x=1}=\ln 2-1.[/tex]
 
  • #8
vanhees71 said:
It's convergent, because the squence [itex]a_n=1/(n+1)[/itex] is monotonically decreasing and going to 0 for [itex]n \rightarrow \infty[/itex]. Then according to the Leibniz criterion the alternating series converges.

To get the value, use the trick of generating functions and define
[tex]f(x)=\sum_{n=1}^{\infty} (-x)^n.[/tex]
This series is convergent for [itex]|x|<1[/itex]. Since it's a geometric series you can sum it explicitly
[tex]f(x)=\frac{1}{x+1}-1.[/tex]
Now you get your series by integrating [itex]f(x)[/itex] from 0 to 1 (argue why!):
[tex]\int_0^1 \mathrm{d} x f(x)=[\ln(x+1)-x]_{x=0}^{x=1}=\ln 2-1.[/tex]

Correct logic, but messes up towards the end and gives the negative of the actual answer. It shouldn't take a genius to see that your answer is negative while the series is supposed to turn out positive.

A way of going with it is to start with the function
[tex]f(x) = \sum_{n=1}^{\infty} x^n = \frac{1}{1-x}-1[/tex]
Integrating this function gives a new function
[tex]F(x) = \sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1} = -\log(1-x)-x+C[/tex]
You can find the constant of integration by setting [itex]x=0[/itex], which gives [itex]C=0[/itex]. Hence, you have:
[tex]\sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1} = -x-\log(1-x)[/tex]
Now set [itex]x=-1[/itex] on both sides to get the answer [itex]1-\log(2)[/itex].

Note that this only finds the value of the series if it actually converges. There are cases in which methods like this are applied to divergent series to find a "value" for them. This is called regularization. To show that the series actually has this value, you would first need to prove that it is convergent.
 

1. What is an alternating series?

An alternating series is a series where the signs of the terms alternate between positive and negative.

2. How do you determine if an alternating series converges or diverges?

To determine if an alternating series converges or diverges, you can use the Alternating Series Test. This test states that if the absolute value of the terms of the series decrease as n increases and approaches 0 as n approaches infinity, then the series converges. If this condition is not met, the series diverges.

3. Can an alternating series converge even if the absolute value of the terms does not approach 0?

Yes, an alternating series can converge even if the absolute value of the terms does not approach 0. This can happen if the series satisfies the Leibniz Criteria, which states that if the terms of an alternating series decrease in absolute value and eventually approach 0, then the series converges.

4. What happens when something disappears in an alternating series?

When something disappears in an alternating series, it means that one or more terms in the series become equal to 0. This can affect the convergence or divergence of the series, depending on the pattern of the terms before and after the disappearance.

5. How can we use alternating series to approximate values?

Alternating series can be used to approximate values by using the Alternating Series Estimation Theorem. This theorem states that the error of an alternating series is less than or equal to the absolute value of the first neglected term. By calculating the sum of a few terms and using the estimation theorem, we can approximate the value of the series to a desired degree of accuracy.

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