Sum of Infinite Series: Finding the Value of S

[docnet]

Homework Statement

.

Relevant Equations

.

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.

 

[Frabjous]

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.

 

[docnet]

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.

 

[Frabjous]

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

 

[docnet]

In regards to the first part, you are almost there. You need to utilize the fact that Σx_k converges.

$\sum x_k$ converges to a sum and $\sum x_k = \sum y_n$, so the latter converges.

For the second part, while I have not done it, I assume that they want you to generate and find the limit of Σy_k for x_k=±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 $\frac{1}{1+1}=\frac{1}{2}$.

 

[Frabjous]

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 x_k 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 x_k∈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 y_k it is the sequence of averages of the partial sums of x_k
5) Since x_k converges, you can get the partial sums of x_k arbitrarily close to the limit so you can also get the averages arbitrarily close although it may take you more terms.

 

[docnet]

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 x_k 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.

2) Regardless, you can only rearrange terms in a series and preserve the limit in general if it is absolutely convergent. Since the x_k∈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..

 

[docnet]

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

 

[docnet]

3) I believe that this problem is being used to motivate Cesaro Sums.
4) If you look at the sequence of partial sums of y_k it is the sequence of averages of the partial sums of x_k
5) Since x_k converges, you can get the partial sums of x_k 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 $$\sigma_n \longrightarrow \frac{1}{2}$$

 

[Frabjous]

The partial sums of the y_k are equivalent to Cesaro partial sums of x_k 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 (p_k) converges, then the sequence ($\sum_{n = 1}^k \frac{p_n}{k}$) converges to the same value. The second part is to use y_k to calculate the "limit".

 

[docnet]

The partial sums of the y_k are equivalent to Cesaro partial sums of x_k 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 (p_k) converges, then the sequence ($\sum_{n = 1}^k \frac{p_n}{k}$) converges to the same value. The second part is to use y_k 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.