# check convergence of a sequence

Pointwise and uniform convergence of $\: f_n(x) = x + \frac{\sin(nx)}n$. Over R? Here's what I've done, I'm not sure if it's correct?

$f_n(x) = x + \frac{\sin(nx)}n$. Over R, c)Fix $x\in R$, $\lim_{x\rightarrow\infty} f_n(x) = \lim_{x\rightarrow\infty}(x + sin(nx)/n)$=x for all fixed x. so $f_n \rightarrow f$ and $f(x)= x$ so it converges pointwise for all x$\in R$. Now for uniform convergence I need to show $\|f_n-f\|_{sup} = sup_R |x+sin(nx)/n -x| = sup_R | sin(nx)/n| \rightarrow\infty$ as $n\rightarrow0$. So it's uniformly convergent.

• It's correct... Commented Feb 25, 2018 at 0:55

I'm not sure what was the question, so I'm supposing that you are asking to review the solution you gave to the first sentence of your question.

The idea of your arguments seems to be correct. But I think your writing could be improved. For example:

• "$f_n\to f$ and $f(x)=x$ so it converges pointwise for all $x\in R$", I would rather prefer writing "so $f_n(x)\to x$, as $n\to\infty$, hence the sequence $(f_n)$ converges pointwise to the function $f:R\to R$, defined by $f(x)=x$."
• "Pointwise and uniform convergence of $f_n(x)=x+\frac{\sin(nx)}x$", could be stated as "Determine the pointwise and uniform convergence of $f_n(x)=x+\frac{\sin(nx)}x$". My English is not that good, so I will not write other suggestions. But keep in mind that it is always a good idea to be precise and write as much as necessary.
• It is worth mentioning that $|\sin(nx)|\le 1$, for all $x\in R$ and all $n\in\mathbb{N}$, since this is a key fact used in your arguments.
• minor detail: there is a typo, it is written $\lim_{x\to\infty}$, but it should be $\lim_{n\to\infty}$.
• I'm just not totally sure if the sequence of functions given is uniformly convergent on R. It is uniformly convergent right?
– Jack
Commented Feb 25, 2018 at 0:48
• @Jack, Yes, by your argument, the function will be uniformly convergent in $R$.
– FYY
Commented Feb 25, 2018 at 0:51