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$$\sum_{n=1}^\infty\frac{(-1)^n}{\sqrt{n}}$$

I did:

$$\lim_{n\to \infty}\Biggr\vert\frac{(-1)^n}{\sqrt{n}}\Biggr\vert$$ $$\lim_{n\to \infty}\frac{1}{n^\frac{1}{2}}=0<1$$

So diverges by the Ratio Test, right?

And have this series:

$$\sum_{n=1}^\infty \bigg(\frac{3n^3+2}{2n^4+1}\bigg)^n$$

I solved and found:

$$\lim_{n\to \infty}\frac{n^4(\frac{3}{n}+\frac{2}{n^4})}{n^4(2+\frac{1}{n^4})}=0<1$$

So is absolutely convergent by the Root Test, right?

But the answers options are:

A)Conditionally convergent and absolutely convergent.

B)Both are absolutely convergent.

C)Both are divergent.

D)Conditionally convergent and divergent.

None matches with my answers. Where am I going wrong?

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3 Answers 3

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Applying the ratio test to the first series is about computing the limit$$\lim_{n\to\infty}\frac{\left\lvert\frac{(-1)^{n+1}}{\sqrt{n+1}}\right\rvert}{\left\lvert\frac{(-1)^n}{\sqrt n}\right\rvert}=\lim_{n\to\infty}\frac{\sqrt n}{\sqrt{n+1}}.$$But this limit is equal to $1$ and therefore the ratio test is inconclusive. Actually, the series converges conditionaly by the alternate series test, but it doesn't converge absolutely.

You are right about the second series.

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  • $\begingroup$ Muito obrigado, José Carlos. Agora entendi onde eu estava errando. $\endgroup$ Commented May 7, 2019 at 18:20
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It converges by Leibniz' criterion for alternating series ($\frac1{\sqrt n}$ converges to $0$ monotonically), but this series does not converges absolutely.

The root text would be only for absolute convergence, since it is a test for series with positive terms.

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As for your first series, you did not show it diverges, but you can show (compare it with $1/\sqrt{n}$) that the series of the absolute values diverges, i.e. your series is not absolutely convergent. Using Leibniz's criteria for alternating sign series you can show it converges, making it simply (or conditionally) convergent. You are correct with respect to the second series, assuming that your calculations indicate you understood the series is bounded by a convergent geometric series.

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  • $\begingroup$ Technically, OP hasn't shown either. All OP has shown is that $a_n\to 0,$ which is not enough to state whether the series $\sum |a_n|$ converges. $\endgroup$ Commented May 7, 2019 at 18:32
  • $\begingroup$ @ThomasAndrews You are correct. $\endgroup$ Commented May 7, 2019 at 18:36

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