# Sum of Infinite Series: Finding the Value of S

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In summary: I see my mistake now. I have corrected it in the summary. Thank you for pointing it out.2) Yes, you are correct. The series is conditionally convergent.3) That makes sense.4) I see now. The Cesaro sum of xk is the sequence of averages and that is the same sequence as the partial sums of yk.5) That makes sense. Thank you for explaining it further.
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Homework Statement
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Relevant Equations
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Using the given rule for the ##x_n##, write $$\sum_l y_l = x_1 + \frac{1}{2} x_2 + \frac{1}{6} x_2 + \frac{1}{3} x_3 + \frac{1}{12} x_2 + \frac{1}{12} x_3 + \frac{1}{20} x_2 + \cdots + \frac{1}{n} x_n$$
$$= x_1 + \sum_{n=2}^\infty \frac{1}{n(n-1)} x_2 + \sum_{n=3}^\infty \frac{2}{n(n-1)} x_3 + \sum_{n=4}^\infty \frac{3}{n(n-1)} x_4 + \cdots + \sum_{k=n}^\infty \frac{n-1}{k(k-1)} x_n$$

in general, ## \sum_{n=1}^m \frac{1}{n(n-1)} = \sum_l ( \frac{1}{n} - \frac{1}{n+1} ) = 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} + \cdots \frac{1}{m-1} - \frac{1}{m} - \frac{1}{m+1} ## The sum collapses like a telescope and everything is canceled out except for the first and last terms. Its limit as ## m \longrightarrow \infty ## is ##1##. So,

$$= x_1 + x_2 + x_3 + \cdots + x_n = \sum_{k=1}^n x_k$$

The infinite series ## \sum_{n=1}^\infty x_n = 1 - 1 + 1 - 1 + \cdots ## diverges since its value alternates between 0 and 1. But, we write ## \sum_n x_n = \sum_l y_l ## and assume the new series converges to a number ##S##.

##\sum_l y_1 = S = 1 + 1 - 1 + 1 + \cdots##

subtract ##1## from both sides

##S - 1 = -1 + 1 + 1 - + \cdots##
##S -1 = - S##
##S = \frac{1}{2}##

i'm unsure about the part where ##S## is used as a variable for a mere algebra trick. it seems like the result from the first part of the question isn't really used.

What is this for? What you have written is incomplete.

You ended up with a divergent series. There are ways to generalize summation for divergent series, but when those techniques are applicable is a specialized topic.

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caz said:
What is this for? What you have written is incomplete.

You ended up with a divergent series. There are ways to generalize summation for divergent series, but when those techniques are applicable is a specialized topic.
I'm sorry. forgot to include the question.

In regards to the first part, you are almost there. You need to utilize the fact that Σxk converges. For the second part, while I have not done it, I assume that they want you to generate and find the limit of Σyk for xk=±1. Hint: look at the partial sums.

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caz said:
In regards to the first part, you are almost there. You need to utilize the fact that Σxk converges.
##\sum x_k## converges to a sum and ##\sum x_k = \sum y_n##, so the latter converges.

caz said:
For the second part, while I have not done it, I assume that they want you to generate and find the limit of Σyk for xk=±1. Hint: look at the partial sums.

I found a few different approaches here, but could not figure out the method you're describing. could you please drop another hint? By Cesaro Sum: The Cesaro sum is one way to assign a value to a divergent series. It computes the arithmetic mean of the first ##n## partial sums and take its limit as ##n\to \infty##.

The partial sums of the Grandi's series are ##\{1,0,1,0,1,...\}##. The average of the first ##2n-1## terms is ##\frac{n}{2n-1}## and the average of the first ##2n## terms is ##\frac{1}{2}##. So, the sequence of averages is ##\{ 1, \frac{1}{2}, \frac{2}{3}, \frac{1}{2}, \frac{3}{5},...\}##
$$\lim_{n \to \infty} \frac{s_1+s_2+s_3+\cdots + s_n}{n} = \frac{1}{2}$$
The Cesaro sum of Grandi's series is ##1/2##.

By Abel's Theorem: Abel's theorem states the sum of a power series that converges for ##|x|<1## , then the value of the series at ##x=1## is the limit of the series as ##x## tends to ##1## from inside of the interval.

Evaluated at ##x=1##, $$\sum_{k=0}^\infty(-x)^k$$ becomes $$\sum_{n=0}^\infty (-1)^n$$ The first series converges to ##\frac{1}{1+x}## with the radius of convergence of ##|x|<1##. By abel's theorem, the value of the series at ##x=1## is ##\frac{1}{1+1}=\frac{1}{2}##.

I looked at this too quickly the first time. I am not a hardcore math guy, so if someone knows a better approach ...

1) Unless I am missing something ##\sum \frac{1}{n(n-1)} = \sum ( \frac{1}{n-1} - \frac{1}{n} )## and you have not dealt with the fact that you do not always start with n=2 and that there are constants multiplying the xk in the individual sums.
2) Regardless, you can only rearrange terms in a series and preserve the limit in general if it is absolutely convergent. Since the xk∈ℝ, this series could only be conditionally convergent
3) I believe that this problem is being used to motivate Cesaro Sums.
4) If you look at the sequence of partial sums of yk it is the sequence of averages of the partial sums of xk
5) Since xk converges, you can get the partial sums of xk arbitrarily close to the limit so you can also get the averages arbitrarily close although it may take you more terms.

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caz said:
1) Unless I am missing something ##\sum \frac{1}{n(n-1)} = \sum ( \frac{1}{n-1} - \frac{1}{n} )## and you have not dealt with the fact that you do not always start with n=2 and that there are constants multiplying the xk in the individual sums.
Terms that sum from ##k>2##,
$$\sum_{n = k}^m \frac{k-1}{n(n-1)} = \sum_{n = k}^m (k-1) \Big( \frac{1}{n-1} - \frac{1}{n} \Big) = (k-1) \Big( \frac{1}{k-1} + \frac{1}{m-1} - \frac{1}{m} - \frac{1}{m+1} \Big) \longrightarrow 1 \quad \text{as}\quad m \longrightarrow \infty$$
so each constant multiplying the ##k_x## sums to 1.

caz said:
2) Regardless, you can only rearrange terms in a series and preserve the limit in general if it is absolutely convergent. Since the xk∈ℝ, this series could only be conditionally convergent
the problem says given a convergent ##\sum x_k## show that ##\sum x_k = \sum y_k##, so seems like there is no need to think about conditionally convergent..

oops.. I posted early by accident... will follow up soon

caz said:
3) I believe that this problem is being used to motivate Cesaro Sums.
4) If you look at the sequence of partial sums of yk it is the sequence of averages of the partial sums of xk
5) Since xk converges, you can get the partial sums of xk arbitrarily close to the limit so you can also get the averages arbitrarily close although it may take you more terms.
$$\sum x_k = 1-1+1-1+1-1+\cdots$$

$$\text{partial sum of x_k}= \quad \begin{cases} s_i = 1 & \text{if i is odd} \quad \\ s_i = 0 & \text{if i is even} \end{cases}$$

$$\sigma_n = \frac{s_1 + s_2 + \cdots + s_n}{n}= \sum_{i=1}^n y_i$$

##s_n \longrightarrow 1/2## as ##n \longrightarrow \infty## and ##\sigma_n \longrightarrow s## as ##n \longrightarrow \infty##

$$\sigma_n = \begin{cases} 1/2 & \text{if n is odd} \quad \\ \frac{n+1}{2n} & \text{if n is even} \end{cases}$$

$$\lim_{n \rightarrow \infty} \frac{n+1}{2n} = \frac{1}{2}$$
so $$\sigma_n \longrightarrow \frac{1}{2}$$

The partial sums of the yk are equivalent to Cesaro partial sums of xk so this problem motivates Cesaro Sums. This first part of the problem is about showing that a convergent series has the same limit as it’s Cesaro Sum. I think you are incorrect in not thinking about conditional convergence.
edit:

You do not need to say Cesaro anywhere. The first part is to show if the sequence (pk) converges, then the sequence (##\sum_{n = 1}^k \frac{p_n}{k}##) converges to the same value. The second part is to use yk to calculate the "limit".

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caz said:
The partial sums of the yk are equivalent to Cesaro partial sums of xk so this problem motivates Cesaro Sums. This first part of the problem is about showing that a convergent series has the same limit as it’s Cesaro Sum. I think you are incorrect in not thinking about conditional convergence.
edit:
View attachment 299465
You do not need to say Cesaro anywhere. The first part is to show if the sequence (pk) converges, then the sequence (##\sum_{n = 1}^k \frac{p_n}{k}##) converges to the same value. The second part is to use yk to calculate the "limit".
Oh ok.. I undertand. The series converges, so the sequence of the averages of partial sums converges too, I think my professor mentioned it is a Cauchy convergent sequence.

## 1. What is a sum of infinite series?

A sum of infinite series is the sum of an infinite number of terms in a sequence. It is a mathematical concept used to find the value of a series that continues infinitely.

## 2. How do you find the value of a sum of infinite series?

To find the value of a sum of infinite series, you can use various mathematical techniques such as the geometric series formula, the telescoping series method, or the ratio test. These methods involve manipulating the terms in the series to find a pattern and determine the limit of the series.

## 3. What is the difference between a convergent and divergent infinite series?

A convergent infinite series is a series whose sum approaches a finite value as the number of terms increases. On the other hand, a divergent infinite series is a series whose sum either approaches infinity or does not have a definite value as the number of terms increases.

## 4. Can all infinite series be summed?

No, not all infinite series can be summed. Only a specific type of series, called a convergent series, can be summed to a finite value. Divergent series cannot be summed as their sum either approaches infinity or does not have a definite value.

## 5. How is the value of a sum of infinite series used in real life?

The value of a sum of infinite series has various applications in real life, such as in finance, physics, and engineering. For example, it can be used to calculate compound interest, determine the total distance traveled by a moving object, or find the total resistance in an electrical circuit.

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