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 ?

  • 1
    $\begingroup$ You should check the intervals when you sum, this doesn't seem correct at all. $\endgroup$ Aug 12, 2021 at 10:40
  • $\begingroup$ @Maxence1402 I think interval will remain constant $\endgroup$
    – jasmine
    Aug 12, 2021 at 10:51
  • $\begingroup$ @Koro yes $n \in \mathbb{N}$ $\endgroup$
    – jasmine
    Aug 12, 2021 at 10:57
  • $\begingroup$ 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. $\endgroup$ Aug 12, 2021 at 12:46
  • $\begingroup$ BTw I haven't had a lot of experiences, @Maxence1402 Sometimes the easiest thing are the most difficult to come up with $\endgroup$
    – jasmine
    Aug 12, 2021 at 13:07

1 Answer 1



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....?

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

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