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I want to determine if this series converges or diverges: $$\sum\limits_{n=1}^\infty{\frac{3^\frac{1}{n} \sqrt{n}}{2n^2-5}}$$

I tried the Ratio Test at first, and didn't get anywhere with that. I'm thinking I have to use Comparison Test, and compare the test to the series $\sum{\frac{1}{n^2}}$ for convergence? But I wasn't sure how to prove that in this case. Can someone help me out here?

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You may write, as $n \to \infty$, $$\frac{3^\frac{1}{n} \sqrt{n}}{2n^2-5}= \frac{\sqrt{n}}{2n^2}\frac{e^{\large \frac{\ln 3}{n}}}{1-\frac{5}{2n^2}}=\frac{1}{2}\frac{1}{n^{3/2}}\frac{1+\frac{\ln 3}{n}+\mathcal{O}\left(\frac {1}{n^2}\right)}{1-\frac{5}{2n^2}}\sim \frac{1}{2}\frac{1}{n^{3/2}}$$ and your initial series is convergent as is the series $\displaystyle \sum\frac{1}{n^{3/2}}$.

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  • $\begingroup$ What's the symbol next to the $\frac{1}{n^2}$? It just looks like a zero? $\endgroup$ Commented Mar 17, 2015 at 22:58
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    $\begingroup$ @sfgiantsfan19 $u_n=\mathcal{O}\left(\frac {1}{n^2}\right)$ means there exists a constant $C$ such that $\left|u_n\right|\leq \frac{C}{n^2}$. Thanks! $\endgroup$ Commented Mar 17, 2015 at 23:01
  • $\begingroup$ Ahh that makes sense now. And no, thank you! $\endgroup$ Commented Mar 17, 2015 at 23:27
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Hint: Do a limit comparison test with $1/n^{3/2}$ since you have a square root in the numerator and a square in the denominator. All you basically need to show after you factor out powers of $n$ is that $3^{1/n} \to 1$.

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