Using a comparison test, test series for convergence / divergence

Question

Test series for convergence / divergence using a comparison test:

$$\sum_{n=1}^\infty\frac{n+2^n}{n+3^n}$$

According to symbollab website, this series converges. I guess I need to compare it with following geometric series (which converges) $$\sum_{n=1}^\infty(\frac{2}{3})^n$$ as less than or equal to, but I don't know how to get rid of n in numerator / denominator.

That’s just a geometric series (ratio is $\frac{2}{3}$ with first term also $\frac{2}{3}$ — Sina Babaei Zadeh, Feb 10 '20 at 22:48
Similar questions are everywhere: Determine if $\sum_{n=1}^\infty \frac{n+4^n}{n+6^n}$ converges, How to prove that $\sum\limits_{n=1}^{\infty}\frac{n+4^n}{n+6^n}$ converge by basics test (comparison, integral, ratio, root) — YuiTo Cheng, Feb 11 '20 at 03:50
Also: http://math.stackexchange.com/questions/879140/is-the-following-series-converging-or-diverging-sum-n-1-infty-dfracn4, http://math.stackexchange.com/questions/3172779/determine-if-this-series-converges-or-diverges-what-test-to-use, http://math.stackexchange.com/questions/571119/show-the-series-sum-j-1-infty-frac2j-j3j-j-converges, http://math.stackexchange.com/questions/3423578/is-the-series-sum-fracn7nn55-converges-or-diverges, — YuiTo Cheng, Feb 11 '20 at 03:51

score 1 · Accepted Answer · answered Feb 10 '20 at 22:54

Use the fact that\begin{align}\lim_{n\to\infty}\frac{\frac{n+2^n}{n+3^n}}{\left(\frac23\right)^n}&=\lim_{n\to\infty}\frac{(n+2^n)3^n}{(n+3^n)2^n}\\&=\lim_{n\to\infty}\frac{n3^n+6^n}{n2^n+6^n}\\&=\lim_{n\to\infty}\frac{\frac n{2^n}+1}{\frac n{3^n}+1}\\&=\frac11\\&=1.\end{align}

score 1 · Answer 2 · answered Feb 10 '20 at 22:57

1

Hint: It can be proved that $n\in\mathbb Z_{>0}\;\implies\; n\le 2^{n-1}$, also $$\frac{n+2^n}{n+3^n}<\frac{n+2^n}{3^n}$$ Then $$\frac{n+2^n}{n+3^n}<\frac{2^{n-1}+2^n}{3^n}=\frac12\left(\frac23\right)^n+\left(\frac23\right)^n$$

answered Feb 10 '20 at 22:57

Ángel Mario Gallegos

19,882

+1 but why not choose $2^n$ ? Angel your last answer is just $\frac 3 2 (2/3)^n$ – user577215664 Feb 10 '20 at 23:22

score 1 · Answer 3 · answered Feb 10 '20 at 23:31

1

$$\sum_{n=1}^\infty\frac{n+2^n}{n+3^n} \le \sum_{n=1}^\infty\frac{2^n+2^n}{3^n}$$ Since $2^n \ge n $ for $n \in \mathbb{N}$ $$\sum_{n=1}^\infty\frac{n+2^n}{n+3^n} \le 2\sum_{n=1}^\infty \left (\frac{2}{3} \right )^n$$ So the series converge.

answered Feb 10 '20 at 23:31

user577215664

40,625

1

(+1) for the most straightforward way – Mark Viola Feb 10 '20 at 23:36
Thank you Mark ... @MarkViola – user577215664 Feb 10 '20 at 23:37

Using a comparison test, test series for convergence / divergence

3 Answers3