Vtyjkyuk

For each $n in mathbbN$ and $x in mathbbR$ define
$$f_n(x) = frac1n;mathrmif;x=nquadmathrmandquad f_n(x) = 0;mathrmif;xneq n.$$
The series $sum f_n(x)$ is uniformly convergent in $mathbbR$?

For each $x in mathbbR$, $f_n(x) = frac1x$ or $f_n(x) = 0$ for every $n in mathbbN$. 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.

By definition of uniform convergence, You need:
$$forall epsilon >0exists NinmathbbNspace s.t. space forall n>N.forall xin mathbbR : |f_n(x)-f(x)|<epsilon $$

Meaning, $N$ is uniform ,depending only on $epsilon$ and not on $x$.

You can also use Cauchy criterion for uniform convergence:

$$forall epsilon >0exists NinmathbbNspace s.t. space forall n,m>N.forall xin mathbbR : |f_n(x)-f_m(x)|<epsilon $$

Note that both of these criterions are equivalent to proving: $sup_xin mathbbR|f_n(x)-f(x)|to0.$

A way to disprove uniform convergence is to find a series of points $x_nto omegainmathbbRcup+infty,-infty $ such that $|f_n(x_n)-f(x_n)|to aneq0$

Similar rules apply when dealing with series , only this time you should look at $S_n=sum_k=1^nf_k(x)$

For $x in mathbb R$ and $n in mathbb N$ you have

$$leftvert sumlimits_k=n+1^infty f_k(x)rightvert le dfrac1n+1$$

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 $n to infty$.

Claim: $|sum_k=j^m f_n (x)|leq frac 1 j$ for all $x in mathbb R$whenever $1leq j leq m$. 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.

Yes, it converges uniformly to the function$$beginarrayrcccfcolon&mathbb R&longrightarrow&mathbb R\&x&mapsto&begincasesfrac1n&text if $x=n$ for some ninmathbb N\0&text otherwise.endcasesendarray$$Just use the definition of uniform convergence and the fact that, for each $varepsilon>0$, $frac1n<varepsilon$ if $ngg1$.

$f_n (x) rightarrow f(x)=0$ $space asspace n rightarrow space infty$
$Nowspace M_n =Sup|f_n| space forall xin R $.
Then $M_n =0 space forall nin N$