Test the uniform convergency of a series of function

Question

Consider the series of function $$\sum_{n=1}^{\infty}\frac{x}{1+n^2x}.$$

Show that this series of function is NOT uniformly convergent in $[0,1]$.

I know only two methods to show a series of function to NOT uniformly convergent :

(i) Using definition of uniform convergent.

(ii) Showing that term by term integration is not possible.

i.e. showing that$$\int_0^1\sum_{n=1}^{\infty}f_n(x)\,dx\not =\sum_{n=1}^{\infty}\int_0^1f_n(x)\,dx.$$

But in each case we need to find out the finite sum $\sum_{k=1}^{n}f_k(x)$.

I am not able to find out this finite sum.

Please help on it....

Does there any other process or theorem to find that the series is not uniformly convergent ?

Answer 1

You may observe that, with $\displaystyle f_n(x)=\frac{x}{1+n^2x}$, $x \in [0,1]$, we have $$ (f_n(x))'=\frac{1}{(1+n^2x)^2}>0 $$ thus $\displaystyle f_n(x)$ is increasing with $x$: $$ 0<\sup_{[0,1]}f_n(x)=f_n(1)=\frac{1}{1+n^2} \leq \frac{1}{n^2}, $$since $\displaystyle \sum\frac{1}{n^2}$ is convergent, hence your initial series is uniformly convergent in $[0,1]$.

Answer 2

Similar to (ii), the strategy is to find a theorem that must be true if the sum is to converge uniformly, and show that it is not true.

One theorem that is applicable in this case is that the limit function must be continuous (See Remark in http://en.wikipedia.org/wiki/Weierstrass_M-test).

But when x=0 the sum is zero and for any nonzero x it is bounded away from zero.

Answer 3

Not an answer, just doesn't fit in the comments.

Olivier's answer shows that the function is continuous on $[0,1]$.

Here is an Octave plot of the sum of the first 100 terms:

In fact, the function is smooth on any compact subset of $(0,1]$.