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

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 $$0 and $$\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.

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

• 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.
• 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.
• 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. Commented Jan 15, 2020 at 20:50
• You could have just said $0\le a_n \le 2^{-n}.$