# find the $\sum_{1}^n f_k?$ and find the $\sum_{1}^{\infty} f_k?$

Let $$\{f_n\}$$ be a sequence of real nonnegative functions on $$\mathbb{R}$$

Given $$f_n(x)=\begin{cases} 1 \ \text{if x} \in [\frac{1}{n+1} ,\frac{1}{n}] \\ 0 \ \text{if x} \notin [\frac{1}{n+1} ,\frac{1}{n}] \end{cases}$$

My question: find the $$\sum_{1}^n f_k ?$$ and find the $$\sum_{1}^{\infty} f_k?$$

My attempt : Given $$f_n(x)=\begin{cases} 1 \ \text{if x} \in [\frac{1}{n+1} ,\frac{1}{n}] \\ 0 \ \text{if x} \notin [\frac{1}{n+1} ,\frac{1}{n}] \end{cases}$$

$$f_1 + f_2 +....+f_n=\sum f_k=\begin{cases} n \ \text{if x} \in [\frac{1}{n+1} ,\frac{1}{n}] \\ 0 \ \text{if x} \notin [\frac{1}{n+1} ,\frac{1}{n}] \end{cases}$$

Is it true ?

For $$\sum_{1}^{\infty} f_k$$

$$f_1 + f_2 +....+f_n+....=\sum f_k=\begin{cases} \infty \ \text{if x} \in [\frac{1}{n+1} ,\frac{1}{n}] \\ 0 \ \text{if x} \notin [\frac{1}{n+1} ,\frac{1}{n}] \end{cases}$$

Is it true ?

• You should check the intervals when you sum, this doesn't seem correct at all. Aug 12, 2021 at 10:40
• @Maxence1402 I think interval will remain constant Aug 12, 2021 at 10:51
• @Koro yes $n \in \mathbb{N}$ Aug 12, 2021 at 10:57
• It baffles me how one seems to have a lot of experience in maths but struggles to compute this simple sum of functions. Koro's answer should be really helpful. Aug 12, 2021 at 12:46
• BTw I haven't had a lot of experiences, @Maxence1402 Sometimes the easiest thing are the most difficult to come up with Aug 12, 2021 at 13:07

Hint:

Let's take one example: suppose that we fix $$x=\frac \pi 4$$ ($$\lt 0.79$$), then clearly $$\frac 1{1+1}\lt x\lt \frac 11$$ and hence $$x\notin [\frac 1{n+1},\frac 1n]$$ for any $$n\gt 1$$ so $$f_1(x)=1$$ and $$f_n(x)=0$$ for all $$n\gt 1$$.

Let $$x\ne 1/i$$ for any $$i\in \mathbb N$$, then for any $$1\gt x\gt 0$$, we have an $$n\in \mathbb N$$ such that $$\frac 1{n+1}\lt x \lt \frac 1n$$

So $$f_1(x)+f_2(x)+\dots+f_n(x)=0+0+\dots+1=1$$

But if $$x=\frac 1i$$ for some $$i\in \mathbb N$$ then note that there will be two closed intervals (of the form $$[\frac 1{m+1}, \frac 1m]$$) sharing $$x$$ so....?

• that means $\sum{f_n}$ will be $\{0,1,2\}$ @ Koro Am i right ? Aug 12, 2021 at 11:13
• yes, depending on values of $x$. I have added one example. You may have a look at that @jasmine
– Koro
Aug 12, 2021 at 11:20