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I'm not sure I got all of these problems right. I'd really appreciate any sort of feedback.

For which $x \in \mathbb{R}$ do the following series converge?

Problem 1

For $\sum\limits_{n=1}^{\infty}\frac{3n^7x^n}{2n!} = \sum\limits_{n=1}^{\infty}\frac{3n^7}{2n!}x^n$, perform a ratio test: $$\lim\limits_{n\to \infty}\left|\frac{3(n+1)^7}{2(n+1)!} \cdot\frac{2n!}{3n^7}\right| = \lim\limits_{n\to \infty}\left|\frac{(n+1)^7}{(n+1) \cdot n^7}\right|= \lim\limits_{n\to \infty} \left|\frac{(n+1)^6}{n^7}\right|\\= \lim\limits_{n\to \infty}\left|\frac{n^6+6n^5+15n^4+20n^3+15n^2+6n+1}{n^7}\right| = \lim\limits_{n\to \infty} 0 =0$$

$\implies$ The ratio of convergence is therefore $r = \infty$.

$\implies$ The power series absolutely converges for any $x \in\mathbb{R}$. $$\\$$

Problem 2

For $\sum\limits_{n=1}^{\infty}\frac{(x-1)^n}{n}$, we see that the coefficients $a_n$ are formed by the harmonic series $a_n = \frac{1}{n}$.

Note that $\lim\limits_{n \to \infty} a_n = 0$.

$\implies$ The ratio of convergence is therefore $r = \infty$.

$\implies$ The power series absolutely converges for any $x \in\mathbb{R}$. $$\\$$

Problem 3

For, $\sum\limits_{n=1}^{\infty}\frac{(-4nx)^n}{2n^3} = \sum\limits_{n=1}^{\infty}\frac{(-4n)^n}{2n^3}x^n$ , perform a root test: $$\lim\limits_{n\to \infty}\left|\sqrt[n]{\frac{(-4n)^n}{2n^3}}\right| = \lim\limits_{n\to \infty}\left|\frac{-4n}{\sqrt[n]{2n^3}}\right|= \lim\limits_{n\to \infty}\left|\frac{-4n}{\sqrt[n]{2}\sqrt[n]{n}\sqrt[n]{n}\sqrt[n]{n}}\right| = \infty, \text{ since} \lim\limits_{n\to\infty} \sqrt[n]{n}=1$$. $\implies$ The radius of convergence is $r = \infty$.

$\implies$ The power series will only converge for $x = 0$.

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  • $\begingroup$ What about the $x$ in (1)? $\endgroup$ – user142299 May 22 '14 at 2:09
  • $\begingroup$ @NotNotLogical What do you mean exactly? In root or ratio tests, you only analyze the $a_n$ part (that is: the coefficients of the power series) from $\sum\limits_{n = 1}^{\infty}a_n x^n$, as far as I know. $\endgroup$ – chiru May 22 '14 at 2:15
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    $\begingroup$ I think you're in confusion here - the ratio/root tests apply to series such as $$\sum_{k=1}^{\infty}b_k$$ For a power series $b_k = a_k x^k$, so you definitely do need to take the $x$ into account. $\endgroup$ – Ayesha May 22 '14 at 2:23
  • $\begingroup$ @Ayesha I just edited (1) and showed that the power series is exactly of the form that you imply. Am I getting it wrong? $\endgroup$ – chiru May 22 '14 at 2:31
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If x=2 in number 2, does the series converge?

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  • $\begingroup$ Woah, I didn't see that! Thanks for pointing out a hidden harmonic series. $\endgroup$ – chiru May 22 '14 at 2:09
  • $\begingroup$ It seems that (2) is convergent for $x \in I = [0,2)$. Thanks a lot for the tip! $\endgroup$ – chiru May 28 '14 at 17:34

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