There is a series of functions $f_n:(0,1) \rightarrow R$ defined as $f_n(x)=\frac{n}{nx+1}$.

The limit of the series is $f(x) = \lim_{n\rightarrow \infty}f_n(x) = \lim_{n\rightarrow \infty}\frac{1}{x+\frac{1}{n}} = \frac{1}{x}$. So $f_n\rightarrow f$ pointwise on $(0,1)$.

Could you please explain:

1) $f$ isn't bounded on $(0,1)$, because $\lim_{x\rightarrow0} \frac{1}{x} = \infty$. Why are $f_n$ bounded on $(0,1)$ (should we consider $n$ strictly less than $\infty$?

2) Why in general pointwise convergence doesn't preserve boundedness, but uniform convergence does? Is the preservation of boundedness by uniform convergence the reason for the possibility of interchanging limits of uniformly convergent series?

  • 2
    $\begingroup$ Did you try to sketch the graphs of $f_n$ and $f$? That will help a lot $\endgroup$ Feb 9, 2017 at 20:30

1 Answer 1


For a fixed $n$, $$|f_n(x)|\leq n,$$ so $f_n$ is bounded.

Uniform convergence preserves boundedness because if $f_n\to f$ uniformly and you advance enough along the sequence, you can find some $f_n$ with $|f_n(x)-f(x)|<1$ for all $x$. Then $$ |f(x)|<|f_n(x)|+1\leq M_n+1 $$ (where $M_n$ is a constant such that $|f_n(x)|<M_n$ for all $x$), and so $f$ is bounded.

And for your last question: yes. Uniform convergence allows you to basically "forget" about the tail of the series and work on finite sums.

  • $\begingroup$ Thank you for the great explanation! If we select $n=\infty$ in your first formula we still have a bounded function? Or we just can't/shouldn't select $n=\infty$ in it? $\endgroup$
    – Konstantin
    Feb 9, 2017 at 20:40
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    $\begingroup$ Infinity is not a number. And even if in some contexts it makes sense to treat it like (almost) a number, you certainly cannot use it to discuss boundedness: if you allow infinity as an upper bound, everything is bounded and so boundedness would mean nothing. $\endgroup$ Feb 9, 2017 at 20:48

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