So I have a question about the convergence of a series here:

It says:

Find the radius and interval of convergence, then identify the values of $x$ for which the series converges absolutely and conditionally

$\displaystyle \sum_{n=1}^{\infty} (\text{csch}(n))x^n$

I already found the radius of convergence. It's $e$. The interval of convergence is $(-e,e)$. I checked the endpoints and they both diverge.

So I know for all values of $x$ on the interval $(-e,e)$, the series converges absolutely. What about the conditional convergence though? How do I show that part? I'm a little confused on that. I don't even think that would ever happen.


The series is not convergent at $x=\pm e$ [ Note that $\frac {e^{n}} {e^{n}-e^{-n}}=\frac 1 {1-e^{-2n}}$ which does not tend to $0$ as $ n \to \infty$. A series $\sum a_n$ cannot converge (even conditionally) unless $a_n \to 0$.

  • $\begingroup$ No. It's definitely $(-e,e)$. You get $|\frac{1}{e}x|<1$ or $|x|<e$. What does it actually mean for a series to conditionally converge? As in, what is the meaning or interpretation of conditional convergence? Is it because at the endpoints, the series is divergent? Suppose it did converge at the endpoints. Because it would then converge at those points, the series would then be conditionally convergent at that point right? $\endgroup$ – Future Math person Feb 26 '20 at 5:45
  • $\begingroup$ @FutureMathperson $\cosh (n)=\frac {e^{n}+e^{-n}} 2$ right? Let $x >0$. Then for $\sum \cosh (n)x^{n}$ to converge both $\sum e^{n}x^{n}$ and $\sum e^{-n}x^{n}$ must converge. This requires both $ex$ and $\frac x e$ to be bounded by $1$. So we must have $x <\frac 1 e$. $\endgroup$ – Kavi Rama Murthy Feb 26 '20 at 5:50
  • $\begingroup$ But my sum isn't $\text{cosh}(x)$. It's $\text{csch(x)}$. $\endgroup$ – Future Math person Feb 26 '20 at 5:52
  • $\begingroup$ @FutureMathperson I have edited my answer. $\endgroup$ – Kavi Rama Murthy Feb 26 '20 at 6:05

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