I want to check the pointwise and uniform convergence of $$\sum_{n=1}^{+\infty}\frac{x^ne^{-n}}{\sqrt{n}}$$

For the pointwise convergence do we check the limit of the sequence?

I mean the following: $$a_n=\frac{x^ne^{-n}}{\sqrt{n}} \rightarrow \lim_{n\rightarrow +\infty}a_n=\lim_{n\rightarrow +\infty}\frac{x^ne^{-n}}{\sqrt{n}}=\lim_{n\rightarrow +\infty}\frac{x^n}{\sqrt{n}e^{n}}=0$$ Therefore the series converges pointwise to $0$.

Is that correct?

And for the uniform convergence do we check also the sequence?

Or do we have to do something else for the series?

  • $\begingroup$ It is not true that if $a_n \to 0$, then $\sum_n a_n$ converges, so your test is insufficient. $\endgroup$ Aug 22 '20 at 13:20
  • $\begingroup$ Uniform convergence mean set for $x$. Which one you have here? $\endgroup$
    – zkutch
    Aug 22 '20 at 13:20
  • $\begingroup$ It might help if you expand on what the definition of "pointwise and uniform convergence" is in this context. When one asks where a power series over $x$ converges pointwise, the question is: for which values of $x$ does it hold that the series converges? $\endgroup$ Aug 22 '20 at 13:23

By well known Cauchy–Hadamard theorem for power series $\sum\limits_{n=0}^{\infty}(z-z_0)^nc_n$ we have, that so called convergence radius $\frac{1}{R}=\lim\limits_{n \to \infty}\sup\sqrt[n]{|c_n|}$. In our case

$$\sqrt[n]{\frac{e^{-n}}{\sqrt{n}}} \to \frac{1}{e}=\frac{1}{R}$$

So we have pointwise convergence for $|x|<e$. In right border point we have divergence as for $\frac{1}{\sqrt{n}}$ and for left convergence $\frac{(-1)^n}{\sqrt{n}}$.

As it is known, uniform convergence we have on each closed segment within convergence interval.

  • 2
    $\begingroup$ @Mary Star. By the way, I wrote for you proof which you asked about on math.stackexchange.com/questions/3797499/… - do you see it? $\endgroup$
    – zkutch
    Aug 22 '20 at 14:02
  • $\begingroup$ For the n-th root test we don't consider the $x^n$-term,right? $\endgroup$
    – Mary Star
    Aug 24 '20 at 5:15
  • $\begingroup$ @Mary Star. Right. Link and formulation added to answer. If/when you'll have more question for this or other answers, feel free to ask. $\endgroup$
    – zkutch
    Aug 24 '20 at 7:46
  • $\begingroup$ @zkutch the ratio accepted answers / answers recently received of this user is tiny! 2 in the last 28 i.e. about 7%. (28 is what's on one page of their account). Seems to be better towards the older of their questions (36 pages!) $\endgroup$ Aug 24 '20 at 11:17
  • $\begingroup$ @Calvin Khor. Thanks. I like that name "Mary".. $\endgroup$
    – zkutch
    Aug 24 '20 at 12:13

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