sadly I don't have any solutions. Is this correct?

$$\sum_{n=1}^\infty\frac{ 2^nx^n}{\sqrt{n^4+1}}$$

We calculate the radius of convergence:

$R=\lim_{x \to\infty}\big|\frac{a_n}{a_{ń+1}}\big|=\lim_{n\to\infty}\frac{1}{2}\sqrt{\frac{(n+1)^4+1}{n^4+1}}=\lim_{n\to\infty}\frac{1}{2}\sqrt{\frac{(1+1/n)^4+1/n^4}{1+1/n^4}}=1/2$

We check the boundary:



By comparing this to the harmonic series, we conclude that it does not converge.


We get


Since $\frac{1}{\sqrt{n^4+1}}\geq 0 \quad \forall n$ and $\lim_{n\to\infty} \frac{1}{\sqrt{n^4+1}} = 0$ and because $\frac{1}{\sqrt{n^4+1}}$ is monotonic decreasing (since the square root is monotonic increasing) we can conclude, using Leibniz-Test, that this series does converge.

So we get: $-\frac{1}{2}\leq x < \frac{1}{2}$

  • $\begingroup$ $\sqrt {n^4+1}\sim n^2$ So that boundary series converges also. Though, as you put $x≤\frac 12$ in the final answer, perhaps it was a typo when you wrote "By comparing this to the harmonic series, we conclude that it does not converge." $\endgroup$ – lulu Jan 27 '19 at 16:44
  • $\begingroup$ Your tests at $|x|=\frac{1}{2}$ are unnecessarily complicated. For large $n$, $\sqrt{n^4+1}\approx n^2$, which immediately gives convergence at both ends of the interval $\endgroup$ – herb steinberg Jan 27 '19 at 16:46
  • $\begingroup$ One may avoid the Leibniz-Test by invoking the absolute convergence of $$\sum_{n=1}^\infty\frac{(-1)^n}{\sqrt{n^4+1}}$$ also it is not 'harmonic' series but you rather mean 'Riemann' series. $\endgroup$ – Olivier Oloa Jan 27 '19 at 16:46
  • $\begingroup$ @lulu the $\leq 1/2$ is a typo. Can you elaborate what exactly you would do with $\sqrt{1+n^4}\sim n^2$? TOlivier: I did mean the harmonic series. $\endgroup$ – xotix Jan 27 '19 at 16:56
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    $\begingroup$ It's an important point. From a computational point of view, mere conditional convergence is generally pretty useless. Theoretically nice, but hard to work with. Absolutely convergent series are much better. $\endgroup$ – lulu Jan 27 '19 at 17:18

The implication is not true for $x={1\over 2}$. By comparison test we have $$\sum_{n=1}^\infty\frac{1}{\sqrt{n^4+1}}<\sum_{n=1}^\infty\frac{1}{\sqrt{n^4}}=\sum_{n=1}^\infty\frac{1}{{n^2}}={\pi^2\over 6}$$


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