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Using the limit comparison test check if the following series converges:

A)

$$a_n=\frac{2^{-n}}{n^2+2n}$$

I take a series $b_n=\frac{2^{-n}}{n^2}$ which converges because $0<b_n<\frac{1}{n^2}$ and $\sum_{n=1}^{\infty}\frac{1}{n^2}<+\infty$.

Since

$$\lim_{n\to \infty}\frac{\frac{2^{-n}}{n^2+2n}}{\frac{2^{-n}}{n^2}}=\lim_{n\to \infty}{\frac{n^2}{n^2+2n}}=1,$$

we conclude that

$$\sum_{n=1}^{\infty}\frac{2^{-n}}{n^2+2n}<+\infty.$$

Is my reasoning correct? It is any way to show this using $b_n$ which is given even in a simpler form?

B)

$$c_n=\frac{3^{n}}{2^n+3^n}$$

I show that the series does not converge by taking $d_n=1^n$, which does not converge, using the fact that

$$\lim_{n\to \infty}\frac{3^{n}}{2^n+3^n}=1.$$

Is it correct?

I would be grateful for any help.

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    $\begingroup$ (A) you could use $b_n=2^{-n}$ instead. (B) Yes. $\endgroup$ Jan 15, 2020 at 19:16

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Yes, it is correct, but it can be made shorter using standard asymptotic equivalence and the rules for asymptotic calculus since you have series with positive terms.

  • For the first series, a polynomial is asymptotically equivalent to its leading term: here, this becomes $n^2+2n\sim_\infty n$, so $$\frac{2^{-n}}{n^2+2n}\sim_\infty \frac{2^{-n}}{n^2}=o\Bigl(\frac1{n^2}\Bigr) ,$$ and the latter is a convergent $p$-series
  • For the second series, $2^n+3^n\sim_\infty 3^n$, hence $$c_n\sim_\infty\frac{3^n}{3^n}=1,$$ so it diverges trivially.
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  • $\begingroup$ I think for the first one it is simpler to keep the numerator rather than the denominator. $0<\frac 1{n^2+n}<1$ and $2^{-n}$ is a CV geometric series. $\endgroup$
    – zwim
    Jan 15, 2020 at 20:46
  • $\begingroup$ I might as well have said that it is $o\Big(\dfrac1{2^n}\Bigr)$. Don't know whether one argument is really simpler than another. I usually use asymptotic analysis because it easily removes irrelevant details. $\endgroup$
    – Bernard
    Jan 15, 2020 at 20:50
  • $\begingroup$ You could have just said $0\le a_n \le 2^{-n}.$ $\endgroup$
    – zhw.
    Jan 15, 2020 at 20:55
  • $\begingroup$ Yes, but I wante d to insist on the usefulness of asymptotic analyisi in these problems. I agree it is not the most convincing example. $\endgroup$
    – Bernard
    Jan 15, 2020 at 21:01

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