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

Question

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 ?

Answer 1

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....?