Test for uniform convergence $\sum\frac{x}{\sqrt{n^3}}$.
Should I define the upper bound $M$ such that $\frac{x}{\sqrt{n^3}}\leq M$, and use the Weierstrass test? But i don’t see if there exist an upper bound. I would appreciate some help. Thanks
Test for uniform convergence $\sum\frac{x}{\sqrt{n^3}}$.
Should I define the upper bound $M$ such that $\frac{x}{\sqrt{n^3}}\leq M$, and use the Weierstrass test? But i don’t see if there exist an upper bound. I would appreciate some help. Thanks
If the series is in fact $\sum\limits_{n=1}^\infty {x\over n^{3/2}} $, then it converges pointwise for all $x$ (since the $p$-series $\sum\limits_{n=1}^\infty {1\over n^p}$ converges for p>1).
But the convergence is not uniform over $\Bbb R$. To see this, note that for any $N$ and $k$ positive integers, we have $ \lim_{x\rightarrow\infty}\ \sum_{n=N}^{N+k} {x\over n^{3/2}} =\infty. $ So, $\sum\limits_{n=1}^\infty {x\over n^{3/2}} $ is not uniformly Cauchy over $\Bbb R$, and thus cannot be uniformly convergent over $\Bbb R$.
If you wanted to determine an interval where the convergence is uniform, the Weierstrass test would be applicable.
To use the test, you first need a set $I$ and a convergent series $\sum\limits_{n=1}^\infty M_n$ of nonnegative terms, such that for each $n$, the inequality $|f_n(x)|\le M_n $ holds for all $x\in I$. Then you would know that the series converges uniformly on $I$.
So, you essentially compare the series of functions with a convergent series of nonegative numbers. Note that you find upper bounds for the terms of the series, not the series itself. Also note that you need to define a set over which the inequalities hold.
A few absolutely convergent series to keep in mind when attempting to apply the test are:
$\ \ \ \ \ $Convergent Geometric series: $\sum\limits_{n=1}^\infty r^n$, $0
$\ \ \ \ \ $Convergent $p$-series: $\sum\limits_{n=1}^\infty {1\over n^p}$, $p>1$.
Evidently, for the series $\sum\limits_{n=1}^\infty {x\over \sqrt{n^3}}=\sum\limits_{n=1}^\infty{x\over n^{3/2}},$ a $p$-series seems applicable.
Indeed $\sum\limits_{n=1}^\infty {1\over n^{3/2}}$ converges, and its terms are nonnegative. The same can be said for $\sum\limits_{n=1}^\infty {D\over n^{3/2}}$, for any constant $D>0$.
Let's check that we have the required inequalities. We choose our set to be the interval $[-D,D]$ for an arbitrary number $D>0$. Then for $|x|\le D$ and for any $n$: $ \Bigl|{x\over\sqrt{n^3}}\Bigr|={|x|\over n^{3/2}}\le {D\over n^{3/2}}. $
It follows that $\sum\limits_{n=1}^\infty {x\over\sqrt{n^3}}$ converges uniformly on $[-D,D]$ for any $D>0$.
If the series is in fact $\sum\limits_{n=1}^\infty {x^n\over n^{3/2}} $, then you could use the ratio test to conclude that it converges for $|x|<1$ and diverges for $|x|>1$. Also, for $|x|=1$, by substitution, it is seen that the series converges.
The convergence would be uniform by the Weierstrass $M$-test. Take $M_n={1\over n^{3/2}}$. Then for $|x|\le 1$: $ \Bigl|{x^n\over\sqrt{n^3}}\Bigr| \le {1\over n^{3/2}}. $ Since $\sum\limits_{n=1}^\infty {1\over n^{3/2}}$ converges, the series $\sum\limits_{n=1}^\infty {x^n\over n^{3/2}} $ converges uniformly on $[-1,1]$.