Series convergence

Sep 2016
15
0
Denmark
Hey,

I have the following two sums:


\(\displaystyle u(x) = \sum_{n=1}^{\infty} \frac{(-1)^{k}}{k} \mathbb{1}_{[k,k+1)}\)

\(\displaystyle u_{n}(x) = \sum_{n=1}^{N} \frac{(-1)^{k}}{k} \mathbb{1}_{[k,k+1)}\)

It sounds trivial but how do I (sufficiently) argue that \(\displaystyle u_{n}(x) \to u(x)\)

My thoughts:

The terms up \(\displaystyle N\) are identical to both sums and letting \(\displaystyle N \to \infty\) we get \(\displaystyle u(x)\).
Or could I simply argue that since \(\displaystyle \vert u(x) - u_{n}(x) \vert \to 0\) then \(\displaystyle u_{n}(x) \to u(x)\).

Is the above sufficient?

Thanks.
 

Walagaster

MHF Helper
Apr 2018
219
143
Tempe, AZ
Hey,

I have the following two sums:


\(\displaystyle u(x) = \sum_{n=1}^{\infty} \frac{(-1)^{k}}{k} \mathbb{1}_{[k,k+1)}\)

\(\displaystyle u_{n}(x) = \sum_{n=1}^{N} \frac{(-1)^{k}}{k} \mathbb{1}_{[k,k+1)}\)
That should be written $u_{n}(x) = \sum_{k=1}^{n} \frac{(-1)^{k}}{k} \mathbb{1}_{[k,k+1)}$

It sounds trivial but how do I (sufficiently) argue that \(\displaystyle u_{n}(x) \to u(x)\)

My thoughts:

The terms up \(\displaystyle N\) are identical to both sums and letting \(\displaystyle N \to \infty\) we get \(\displaystyle u(x)\).
Or could I simply argue that since \(\displaystyle \vert u(x) - u_{n}(x) \vert \to 0\) then \(\displaystyle u_{n}(x) \to u(x)\).

Is the above sufficient?

Thanks.

You have the idea, but just stating it like that wouldn't be adequate to hand in for a proof that $u_n(x) \to u(x)$ as $n \to \infty$. You need an argument like: Suppose $x > 0$. Given $\epsilon > 0$ then show how to find an $N$ such that if $n > N$, $|u_n(x) - u(x)| < \epsilon$.
 

Plato

MHF Helper
Aug 2006
22,462
8,633
I have the following two sums:
\(\displaystyle u(x) = \sum_{n=1}^{\infty} \frac{(-1)^{k}}{k} \mathbb{1}_{[k,k+1)}\)
\(\displaystyle u_{n}(x) = \sum_{n=1}^{N} \frac{(-1)^{k}}{k} \mathbb{1}_{[k,k+1)}\)
It sounds trivial but how do I (sufficiently) argue that \(\displaystyle u_{n}(x) \to u(x)\)
I have a question about notation here. When one sees $u(x)$ that is $u$ is a function of $x$.
However in $u(x) = \sum\limits_{k = 1}^\infty {\dfrac{{{{( - 1)}^k}}}{k}{1_{[k,k + 1)}}}$ there is no $x$ on the RHS.
I assume that it should be the indicator function: ${\large 1}_{[k,k+1)}(x)=\begin{cases} 1 &: x\in[k,k+1) \\0 &: \text{ else }\end{cases}$
Now that really makes things odd. Because$k$ is an index in the sum it is a positive integer.
Thus if $x> 0$ then there is exactly one integer $J = \left\lfloor x \right\rfloor < x < (\left\lfloor x \right\rfloor + 1) = J + 1$
That means $u(x)=\dfrac{(-1)^{\left\lfloor x \right\rfloor}}{\left\lfloor x \right\rfloor}$ for $x>0$ and zero otherwise.

Now consider the sequence $u_n(x)$. If $x<N$ then $u_n(x)=0$ but if $x\ge N$ then $u_n(x)=\dfrac{(-1)^{\left\lfloor x \right\rfloor}}{\left\lfloor x \right\rfloor}$
Perhaps I have grossly misread the function? But in any case I find the an odd question.
 

Walagaster

MHF Helper
Apr 2018
219
143
Tempe, AZ
Yes, the indicator function should have an $x$ variable. Including that, did you notice the correction I made for the OP in post #2 that $u_n(x)$ should be written $u_{n}(x) = \sum_{k=1}^{n} \frac{(-1)^{k}}{k} \mathbb{1}_{[k,k+1]}(x)$? Not sure if that is what is bothering you.
 

Plato

MHF Helper
Aug 2006
22,462
8,633
Yes, the indicator function should have an xx variable. Including that, did you notice the correction I made for the OP in post #2 that un(x)un(x) should be written un(x)=∑nk=1(−1)kk1[k,k+1](x)un(x)=∑k=1n(−1)kk1[k,k+1](x)? Not sure if that is what is bothering you.
Thank you for your reply. Yes I did notice your correction, however I am just bothered by the oddness of the question.
Because ⌊x⌋⩽x<⌊x⌋+1 then x∈[k,k+1)x∈[k,k+1) for no more than exactly one index k.
That means that in the sum $u(x)$ there is only one non-zero term $\dfrac{{{{( - 1)}^{\left\lfloor j \right\rfloor }}}}{{\left\lfloor j \right\rfloor }}$.
where $j$ is an index such that $x\in[j , j+1)$ assuming $x>0$.
Moreover, if $n \geqslant \left\lfloor x \right\rfloor > 0$ then $u_n(x)=\dfrac{{{{( - 1)}^{\left\lfloor x \right\rfloor }}}}{{\left\lfloor x \right\rfloor }}$
 

Walagaster

MHF Helper
Apr 2018
219
143
Tempe, AZ
No, I think you are interpreting it wrong. Each term in the sum is a function defined for all $x$ and is nonzero on the corresponding unit long interval and zero elsewhere. So the sum of the first n terms is a step function which is the sum of the first n characteristic functions times their coefficients. I would have preferred the usual notation $\chi_{[k,k+1)}(x)$ for the characteristic functions.
 

Plato

MHF Helper
Aug 2006
22,462
8,633
No, I think you are interpreting it wrong. Each term in the sum is a function defined for all $x$ and is nonzero on the corresponding unit long interval and zero elsewhere. So the sum of the first n terms is a step function which is the sum of the first n characteristic functions times their coefficients. I would have preferred the usual notation $\chi_{[k,k+1)}(x)$ for the characteristic functions.
Yes I too prefer the chi notation for the characteristic function: $\chi_{[k,k+1)}(x)=\begin{cases}1 &: x\in [k,k+1) \\ 0 &: \text{ else }\end{cases}$
So lets see if you are it interpreting it wrongly. Lets simplify typing $u(x) = \sum\limits_{k = 1}^\infty {{k^2}{\chi _{[k,k + 1)}}(x)} $
Using that simplified function are these true?: $u(-2)=0,~u(2.8)=4~\&~u(10.99)=100$
Now lets bump this up to the sequence: ${u_n}(x) = \sum\limits_{k = 1}^n {{k^2}{\chi _{[k,k + 1)}}(x)} $ (I think that incorporates your correction.)
Are these correct: $\begin{array}{*{20}{c}} {u_9(-2)=0}&{u_9(2.6)=4}&{u_9(10.9)=0} \\ {u_{10}(-2)=0}&{u_{10}(2.8)=4}&{u_{10}(10.9)=100} \\ {u_{11}(-2)=0}&{u_{11}(2.8)=4}&{u_{11}(10.9)=100} \end{array}~?~?$
If they are correct then how have I miss-interrupted the function?
 
Last edited:

Walagaster

MHF Helper
Apr 2018
219
143
Tempe, AZ
Those numbers look correct. In your earlier post you had this statement:

Now consider the sequence $u_n(x)$. If $x<N$ then $u_n(x)=0$ but if $x\ge N$ then $u_n(x)=\dfrac{(-1)^{\left\lfloor x \right\rfloor}}{\left\lfloor x \right\rfloor}$
Perhaps I have grossly misread the function? But in any case I find the an odd question.
Maybe we're getting $n$ and $N$ mixed up by the confusion in the original post. I'm assuming you are using $N$ for the number of terms in the partial sum. You have if $x<N$ then $u_n(x) = 0$. I don't agree. And surely also for $x> N+1,~u_n(x) = 0$
I think we may be talking at cross purposes here. Do you agree that $u_{n}(x) = \sum_{k=1}^{n} \frac{(-1)^{k}}{k} \chi_{[k,k+1]}(x) $ is a step function which is $0$ for $x \ge n+1$?