# Series convergence

#### Meelas

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
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
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
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
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
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
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?

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#### Walagaster

MHF Helper
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 $x \ge n+1$?