I have these (math) problems where I am first supposed to find out if a given series converge or diverge, and then, by using that result, multiplying the same series by $x^n$ and find out for which values of $x$ the series converge absolutely, conditionally or diverge. I think I am starting to get a certain idea of which converge/divergence tests to use for different types of series, but the "by using that result"-part is confusing me a little bit.

First series

I concluded $\sum_{n=2}^{\infty}\frac{n^2 + 1}{n^2 - 1}$ diverges by applying the divergence test: $\lim_{n \to \infty}\frac{n^2 + 1}{n^2 - 1} = 1 \neq 0$.

However, when I try to find out for which values of $x$ the series $\sum_{n=2}^{\infty}\frac{n^2 + 1}{n^2 - 1}x^n$ converge absolutely, conditionally or diverge, I dont know how to directly use the result I found previously. My book says the way to find out the radius of converge is by using the ratio test:

$ \frac{1}{R} = L = \lim_{n \to \infty}\left|\frac{a_{n + 1}}{a_n}\right| = \lim_{n \to \infty}\frac{((n+1)^2 +1)(n^2 - 1)}{((n+1)^2 - 1)(n^2 + 1)} = 1$ (by multiplying by $\frac{1}{n^2}$)

So because radius of convergence is $1$, I concluded the series converges absolutely for $x \in (-1 , 1)$ and diverge for $(-\infty, -1]\cup[1, \infty)$. Here I knew already that the series diverges for $x = 1$ and thus can not converge absolutely for $x \leq -1$. It does also not converge conditionally as $\lim_{n \to \infty}\frac{n^2 + 1}{n^2 - 1}(-1)^n \neq 0$.

So, in other words, assuming I am on the right track here, I need a little advice how to do this more direct if possible. I do one more example to see if I have thought about it in the right way..

Second series

I used the comparison test to conclude that $\sum_{n=0}^{\infty}\frac{2^n}{4^n + 1}$ converges since $\frac{2^n}{4^n + 1} \lt \frac{1}{2^n}$ for $n = 0, 1, 2, \dots$

Now, trying to do the same as previously for $\sum_{n=0}^{\infty}\frac{2^n}{4^n + 1}x^n$, I don't see a way to use what I just found directly, so I do the ratio test again to find the radius of converge with center of converge at $x = 0$:

$\frac{1}{R} = L = \lim_{n \to \infty}|\frac{a_{n + 1}}{a_n}| = \lim_{n \to \infty}\frac{2^{n + 1}(4^n + 1)}{2^n(4^{n + 1} + 1)} = \lim_{n \to \infty}\frac{2*4^n}{4*4^n + 1} + \lim_{n \to \infty}\frac{2}{4*4^n + 1} = \frac{2}{4} \Rightarrow R = 2$. So I concluded the series converges (absolutely) for $x \in (-2, 2)$ which actually makes sense when I look at it, as $x = 2$ would make the numerator $4^n$ and the series would diverge like $\sum_{n=0}^{\infty}1$. Again I do not see any way to make the series converge conditionally and thus diverges everywhere else but $x \in (-2, 2)$


I need advice how I can use the information about whether a series converges or diverges to decide for which values of the same series multiplied by $x^n$ converges absolutely, converges conditionally or diverges. Any help appreciated, sorry if the math things are a little bit tiny.


I don't think there's really a better way to do either of these. A slightly less direct but nicer way would be to use the limit comparison test to compare it with $$\sum_{n=2}^{\infty} x^n$$ (for the first series) or $$\sum_{n=2}^{\infty} (x/2)^n$$(for the second), but it's essentially the same idea. Generally speaking, if you can determine how your original series converges, you can find the radius convergence of it multiplied by $x^n$ in this manner.

(Also, a note: If you put double-dollar-signs around expressions they center and become bigger - that's called displaymath mode and it's generally used for expressions with large operators.)

  • $\begingroup$ But if i use the comparison test, how do I know how the original series converges? My book says if I have series $\{a_n\}$ and $\{b_n\}$ then I know that $\{a_n\}$ converges if $L < \infty$ and $\{b_n\}$ converges or diverges if $L > 0$ and $\{b_n\}$ diverges where $L = \lim_{n \to \infty}\frac{a_n}{b_n}$. It says nothing about <b>how</b> $\endgroup$ – Amoz Feb 21 '18 at 12:02
  • $\begingroup$ I am very curious because I am trying to do do the same thing for $$\sum_{n=1}^{\infty} \frac{n}{n^{2}\sqrt{n} + 1}$$ which wasn't so easy to do with the ratio test, but I know it behaves like $$\frac{1}{n^{\frac{3}{2}}}$$ $\endgroup$ – Amoz Feb 21 '18 at 16:50
  • 1
    $\begingroup$ @Amoz The idea is that, if $L=\lim_{n\to\infty} \frac{a_n}{b_n}$ is a positive real number (i.e. $0<L<\infty$), then the series $\sum_{n=1}^{\infty} a_n$ converges iff the series $\sum_{n=1}^{\infty} b_n$ does. In other words, if their ratio tends to a nonzero finite constant, the two series behave the same way. It's usually used where one can determine whether $\sum_{n=1}^{\infty} b_n$ converges relatively simply. $\endgroup$ – Carl Schildkraut Feb 21 '18 at 21:27

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