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$$ \sum_{k=0}^{\infty} \left( \frac{k^2+1}{2k^2 - 1} \right)^k sin(2k) $$

What is the best approach that we should have to check for absolute convergence in the following series?

Here's what I've tried:

Applying the root test we'll get:

$$ \lim_{k\rightarrow\infty}\sqrt[k]{\left| \left( \frac{k^2+1}{2k^2 - 1} \right)^k sin(2k) \right|}\\ \lim_{k\rightarrow\infty}\left|\frac{k^2+1}{2k^2 - 1} \right| \sqrt[k]{ \left| sin(2k) \right|}\\ $$

Ignoring the sine, the fraction as $k \rightarrow \infty$ will approach $\frac{1}{2}$, which is $ < 1$ and because of that will absolutely converge. But can I ignore the sine? We're analyzing its absolute value... So we'll have $0 \leq sin(2k)\leq 1$. So as $k \rightarrow \infty$ the sine is not going to oscillate in signs... And that's why I've ignored the sine, but I don't know if I'm able to do that.

Can someone please explain to me the best way to get the correct answer?

Thanks!

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    Hint: Clearly $0<\sqrt[k]{|sin(2k)|}<1$ and it is multiplied by $1/2$ from the fraction – 2017-02-08
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    @Hugh isn't $0\leq\sqrt[k]{|sin(2k)|}\leq1$ ? if so, we'll get that its value is not going to make the limit $> 1$ so it's going to be absolutely convergent. Is that?! – 2017-02-08
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    it works that way too, although you can tighten the bounds to $0<\cdot<1$ because $y=sin(x)$ requires an irrational $x$ to get $y=0$ or $y=1$ – 2017-02-08
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    @Hugh Thank you my friend. Awesome explanation... I had the feeling that I was able to do that but couldn't figure it out on why... Thanks – 2017-02-08

3 Answers 3

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In fact, the root test criterion relies on the computation of $\limsup a_k$, which always exists in $\mathbb{R} \cup \{\infty\}$ (if you use only $\lim$, you have to check that the limit really exists when you majorate). In that case, the majoration $|\sin| \le 1$ can be applied:

$$\limsup_{k \to \infty} \left|\left(\frac{k^2+1}{2k^2-1}\right)^k\sin(2k)\right|^{\frac{1}{k}} \le \limsup_{k \to \infty} \left|\frac{k^2+1}{2k^2-1}\right|\times 1^\frac{1}{k} \to \frac{1}{2}$$

Actually you can even state directly that:

$$\left|\left(\frac{k^2+1}{2k^2-1}\right)^k\sin(2k)\right| \le \left|\frac{k^2+1}{2k^2-1} \right|^k$$

And as you showed $\sum \left|\frac{k^2+1}{2k^2-1} \right|^k$ converges, then by majoration $\sum \left|\left(\frac{k^2+1}{2k^2-1}\right)^k\sin(2k)\right|$ converges.

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    Can I use the fact that the value of sine is going to be $\leq 1$ and because of the limit of my fraction is $\frac{1}{2}$ it'll never be $> 1$ hence always absolutely convergent?! – 2017-02-08
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    @BrunoReis: That's the main idea of the proof. – 2017-02-08
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It doesn't matter. For example, for any $\epsilon>0$ you have $$\sqrt[k]{a_{k}}<\frac{1}{2}+\epsilon$$ for sufficiently large $k$, so the series converges.

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    Thank you for answering... I don't think I get it... Can you explain it better please? – 2017-02-08
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Yes you are allowed to do that and it is valid. But you can get the result by completely ignoring sine term and that summation converges, and all of its terms are positive. Now you are multiplying sine function to each term, so a few terms will be less than original terms. So the final function also converges.