# Uniform convergence of the series of functions $\sum\frac{1}{n+n^2x}$ on $(0,1]$.

How to check uniform convergence of the series of functions $$\sum\frac{1}{n+n^2x}$$ on the interval $$(0,1]?$$ I tried it by various methods but didn’t find any suitable way. $$M_n$$- test , Abel’s and Dirchlet test does not give appropriate answer . I tried by taking derivative series also . Please suggest how to check it’s uniform convergence. Thanks in advance.

• I assume you are talking about the functions $f_N : (0,1] \to \mathbb R$ defined by $f_N(x) = \sum^N_{n=1} \frac{1}{n+n^2x}$ for $x \in (0,1]$, $N \in \mathbb N$, and you want the convegence to be uniform in $x$ as $N \to \infty$. Unfortunately, the convergence is not uniform. This can be seen because $f_N$ is bounded for each $N$, and if the convergence was uniform this means that the limiting function $f$ would need to be bounded too. Here the limiting function is unbounded. – User8128 Sep 27 '19 at 16:04
• @User8128 what is limiting function ... – neelkanth Sep 27 '19 at 16:08
• Thanks for reply ... – neelkanth Sep 27 '19 at 16:08

The only possible issues are in a right neighbourhood of the origin: if we assume $$\frac{1}{m+1}\leq x\leq \frac{1}{m}$$ we have $$-\log(x)\leq\log(m+1)\leq H_m=\sum_{n\geq 1}\frac{m}{n(n+m)} \leq \sum_{n\geq 1}\frac{1}{n+n^2 x}=f(x)$$ hence the converge is not uniform, since $$f(x)$$ is unbounded in a right neighbourhood of the origin.

If the series converged uniformly, then we would have

$$\lim_{n \to \infty} \sup _{x \in (0,1]}\left|\sum_{k = n+1}^{\infty}\frac{1}{k + k^2x}\right| = 0$$

This is a direct consequence of the definition that $$f_n(x) \to f(x)$$ uniformly for $$x \in D$$, if for any $$\epsilon > 0$$ there exists $$N \in \mathbb{N}$$ independent of $$x$$ such that $$|f_n(x) - f(x)| < \epsilon$$ for all $$n > N$$and all $$x \in D$$.

However, since $$x_n = 1/n \in (0,1]$$,

$$\left|\sup _{x \in (0,1]}\sum_{k = n+1}^{\infty}\frac{1}{k + k^2x}\right|\geqslant\sup _{x \in (0,1]}\sum_{k = n+1}^{2n}\frac{1}{k + k^2x} \geqslant \sup _{x \in (0,1]} \frac{n}{2n + 4n^2x} \\\geqslant \frac{n}{2n + 4n^2(1/n)} = 1/6 \not\to 0$$

Thus, convergence is not uniform on $$(0,1]$$.

If $$f_N(x)=\sum_{n=1}^N \frac{1}{n+n^2x}$$ converged uniformly, the sequence $$\sum_{n=1}^N \frac{1}{n}$$ should converge, which is absurd.

• @lzrabu ininterval zero in not included... then why to talk about $\sum 1/n$ – neelkanth Sep 28 '19 at 2:10
• Since $0$ is a limit point of $(0, 1]$, the uniform convergence of $f_n$ let us conclude the convergence of $\lim_{x \rightarrow 0^+} f_N(x)$. – Lázaro Albuquerque Sep 28 '19 at 3:58