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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

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    What does ∑x|√n3 mean?2012-01-06
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    In its present state, this question is illegible.2012-01-06
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    @neemy Do you perhaps mean $$\sum\limits_{n=1}^\infty \frac{x}{\sqrt{n^3}}$$ because as it stands we have no idea. If you clarify your question we may be able to help.2012-01-06
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    Yes that is the meaning!2012-01-06
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    @neemy: Could you please specify the domain? Without it, it is impossible to know what you mean by uniform convergence.2012-01-06
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    @neemy: Please learn the basics of LaTeX if you want to write your questions legibly.2012-01-06
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    Even if you must use ASCII, at least make it unambiguous. Even pseudocode like `sum(n=1 to infinity, x/sqrt(n^3))` is better.2012-01-06
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    using D Alembert test i found that the series converges when |x|<1 and diverges when |x|>1.Does this imply it is uniform convergence?2012-01-06
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    Uh, since you mentioned the ratio test, are you sure you don't mean $$\sum_{n=1}^{\infty} \frac{x^n}{\sqrt{n^3}}$$ instead?2012-01-06

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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 $[-1,1]$.