How to prove that $\sum\limits_{n=1}^{\infty}\frac{n+4^n}{n+6^n}$ converge by basics test (comparison, integral, ratio, root)

Question

I have problem to prove that this series is converge. I know that it's converge without any proof but don't know how to prove it. $$\sum_{n=1}^{\infty}\frac{n+4^n}{n+6^n}$$

$$0\leqslant\frac{n+4^n}{n+6^n}\leqslant\frac{4^n+4^n}{0+6^n}=2\cdot\left(\frac{2}{3}\right)^n$$ — Did, May 02 '15 at 10:56

score 3 · Accepted Answer · answered May 02 '15 at 11:01

3

$$\sum_{n=1}^\infty\frac{n+4^n}{n+6^n}<\sum_{n=1}^\infty\frac{5^n}{6^n}=\frac56\cdot\frac1{1-\frac56}=5$$

since $$4^n+n<(4+1)^n$$

answered May 02 '15 at 11:01

ajotatxe

65,084

score 1 · Answer 2 · answered May 02 '15 at 10:51

1

For instance: $$ \left(\frac{n+4^n}{n+6^n}\right)^{1/n} \to \frac 46 \in (-1,1) $$

answered May 02 '15 at 10:51

mookid

28,236

The root test will work like that if $a_n = \frac{4^n}{6^n}$ but here it's a little different so I can't use the root test – aukxn May 02 '15 at 10:55
of course you can. why not? – mookid May 02 '15 at 10:55

Frank Newman · Answer 3 · 2015-05-02T17:30:34.580

0

Whenever I see addition of a strong term plus a weak term, I consider the limit comparison test. Here, we have strong plus weak in both the numerator and the denominator. Choose the new sequence to be the fraction with only the strong parts included: $\frac{4^n}{6^n}$. Read up on the limit comparison test if you're not sure how to proceed.

edited May 02 '15 at 17:30

answered May 02 '15 at 16:28

Frank Newman

311

How to prove that $\sum\limits_{n=1}^{\infty}\frac{n+4^n}{n+6^n}$ converge by basics test (comparison, integral, ratio, root)

3 Answers3