Consider the series $$\sum_{n=1}^{\infty}u_n(x)$$ for which the sum of first n terms $$S_n(x)=\frac{\log(1+n^4x^2)}{2n^2}, x \in [0,1]$$.Prove that the series of derivatives $$\sum_{n=1}^{\infty}u'_n(x)$$ does not converge uniformly on $x \in [0,1]$.

What I have in hand that is $$S_n(x)=\frac{\log(1+n^4x^2)}{2n^2}, x \in [0,1]$$.
Now breaking and extending the terms I have $$\sum_{n=1}^{\infty}u_n(x)=\sum_{n=1}^{\infty}\frac{\log(1+n^4x^2)}{2n^2}=\frac{\log(1+x^2)}{2}+\frac{\log(1+2^4x^2)}{2.2^2}+....+\frac{\log(1+({n-1})^4x^2)}{2.({n-1})^2}+\frac{\log(1+n^4x^2)}{2.n^2}+...+\infty$$

$$\Rightarrow \sum_{n=1}^{\infty}u'_n(x)=\sum_{n=1}^{\infty}\frac{n^2.x}{1+n^4x^2}=\frac{x}{1+x^2}+\frac{2^2.x}{1+2^4x^2}+\frac{3^2.x}{1+3^4x^2}+...$$

Now by ratio Test I have,
$$\lim_{n\to\infty}\left|\frac{u'_{n+1}(x)}{u'_n(x)}\right|=\lim_{n\to\infty}\left|\frac{(n+1)^2.x}{1+(n+1)^4x^2}.\frac{1+n^4x^2}{n^2.x}\right|\rightarrow1 \neq0$$

Is this the reason behind non-uniform convergence?

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  • $\begingroup$ If the ratio test goes to $1$, it doesn't mean the series diverges. It just means that the test can say nothing about convergence. $\endgroup$ – Alfred Yerger Dec 24 '17 at 15:58
  • $\begingroup$ If the series diverges then there is no meaning of uniform convergence. So from this part is my attempt correct to prove that it is not uniform? $\endgroup$ – vbm Dec 24 '17 at 16:05
  • $\begingroup$ But the series does not diverge. It is pointwise convergent. To see this, note that $1 + n^4x^2 > n^4 x^2$, so the terms are bounded by $1/n^2x$ which is pointwise convergent. $\endgroup$ – Alfred Yerger Dec 24 '17 at 16:06
  • $\begingroup$ One more specific confusion: - if the terms are bounded by $1/n^2x$ , then by p-test for $n>1$ doesn't it imply the uniform convergence instead of pointwise convergence? $\endgroup$ – vbm Dec 24 '17 at 16:10
  • $\begingroup$ The p test only implies pointwise convergence, since for different $x$ you get radically different summations. Think about what happens as $x \to 0$! $\endgroup$ – Alfred Yerger Dec 24 '17 at 16:11

Notice that the terms go to $0$ pointwise. If convergence was uniform, given an $\epsilon$, we should be able to find an $N$ so that for $n > N$, all the terms are uniformly small, by the Cauchy criterion for uniform convergence. However, taking the point $x = 1/n^2$, the summand evaluates to $2$, so there is always at least one point which violates the criterion, so convergence is not uniform.

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