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.
$\begingroup$
$\endgroup$
3
-
$\begingroup$ 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. $\endgroup$ – User8128 Sep 27 '19 at 16:04
-
$\begingroup$ @User8128 what is limiting function ... $\endgroup$ – neelkanth Sep 27 '19 at 16:08
-
$\begingroup$ Thanks for reply ... $\endgroup$ – neelkanth Sep 27 '19 at 16:08
Add a comment
|
$\begingroup$
$\endgroup$
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.
answered Sep 27 '19 at 16:09
$\begingroup$
$\endgroup$
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 $(0,1]$.
$\begingroup$
$\endgroup$
2
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.
answered Sep 27 '19 at 18:23
-
$\begingroup$ @lzrabu ininterval zero in not included... then why to talk about $\sum 1/n$ $\endgroup$ – neelkanth Sep 28 '19 at 2:10
-
$\begingroup$ Since $0$ is a limit point of $(0, 1]$, the uniform convergence of $f_n$ let us conclude the convergence of $\lim_{x \rightarrow 0^+} f_N(x)$. $\endgroup$ – Lázaro Albuquerque Sep 28 '19 at 3:58
