$f_{n}(x) = \frac{1}{n}$ if $x=n$ and $f_{n}(x) = 0$ if $x\neq n$. Is $\sum f_{n}(x)$ uniformly convergent? 
For each $n \in \mathbb{N}$ and $x \in \mathbb{R}$ define
  $$f_{n}(x) = \frac{1}{n}\;\mathrm{if}\;x=n\quad\mathrm{and}\quad f_{n}(x) = 0\;\mathrm{if}\;x\neq n.$$
  The series $\sum f_{n}(x)$ is uniformly convergent in $\mathbb{R}$?

For each $x \in \mathbb{R}$, $f_{n}(x) = \frac{1}{x}$ or $f_{n}(x) = 0$ for every $n \in \mathbb{N}$. So, the series converges (pointwise). Now, about uniformly convergence, I'm confuse. I don't see the criterion to apply in this series. I appreciate any hints. 
 A: By definition of uniform convergence, You need: 
$$\forall \epsilon >0\exists N\in\mathbb{N}\space s.t. \space \forall n>N.\forall x\in \mathbb{R} : |f_n(x)-f(x)|<\epsilon $$
Meaning, $N$ is uniform ,depending only on $\epsilon$ and not on $x$.
You can also use Cauchy criterion for uniform convergence:
$$\forall \epsilon >0\exists N\in\mathbb{N}\space s.t. \space \forall n,m>N.\forall x\in \mathbb{R} : |f_n(x)-f_m(x)|<\epsilon $$
Note that both of these criterions are equivalent to proving: $\sup_{x\in \mathbb{R}}|f_n(x)-f(x)|\to0.$
A way to disprove uniform convergence is to find a series of points $x_n\to \omega\in\mathbb{R}\cup\{+\infty,-\infty\} $ such that $|f_n(x_n)-f(x_n)|\to a\neq0$
Similar rules apply when dealing with series , only this time you should look at $S_n=\sum_{k=1}^nf_k(x)$
A: For $x \in \mathbb R$ and $n \in \mathbb N$ you have 
$$\left\vert \sum\limits_{k=n+1}^\infty f_k(x)\right\vert \le \dfrac{1}{n+1}$$
Therefore $\sum f_n$ converges uniformly on $\mathbb R$: the supremum of the reminder $R_n(x)$ is uniformly bounded by a quantity which vanishes as $n \to \infty$.
A: Claim: $|\sum_{k=j}^{m} f_n (x)|\leq \frac 1 j$ for all $x \in \mathbb R$whenever $1\leq j \leq m$. Once we show this uniform convergnc e follows. Fxi $x$. If $x$ is not in  $\{j,j+1,...,m\}$ then  $\sum_{k=j}^{m} f_n (x)=0$. If $x=k$ for some $k$ in this range then the sum is exactly $\frac 1 k$ which does not exceed$\frac 1 j$. This proves teh claim. 
A: Yes, it converges uniformly to the function$$\begin{array}{rccc}f\colon&\mathbb R&\longrightarrow&\mathbb R\\&x&\mapsto&\begin{cases}\frac1n&\text{ if $x=n$ for some }n\in\mathbb N\\0&\text{ otherwise.}\end{cases}\end{array}$$Just use the definition of uniform convergence and the fact that, for each $\varepsilon>0$, $\frac1n<\varepsilon$ if $n\gg1$.
A: $f_{n} (x) \rightarrow f(x)=0$ $\space as\space n \rightarrow \space \infty$
$Now\space M_{n} =Sup|f_{n}| \space \forall x\in R $. 
Then $M_{n} =0 \space \forall n\in N$
