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I know one solution.

That is by Doing comparison with $\dfrac{4^n+4^n}{6^n}$

Wondering if there are more ways to do it

Holy cow
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  • Wolfram Alpha gives the series as convergent to $S\approx2.0897$. – Jam Jul 27 '14 at 00:09
  • when the size of matter increases ppl use ratios, but then its not the right way to solve problems because they increase the density of fluids, so life is tough – Holy cow Jul 27 '14 at 00:29
  • You could also use the Limit Comparison Test with $\sum\frac{4^n}{6^n}$. – user84413 Jul 27 '14 at 00:37

3 Answers3

2

Comparison is good. Most students would automatically go to the Ratio Test. That also works well, but to me is less attractive.

André Nicolas
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1

By the ratio test, we have $$ r=\lim_{n \to +\infty}\biggl|\frac{a_{n+1}}{a_n}\biggr| = \lim_{n \to +\infty}\frac{n+1 + 4^{n+1}}{n + 4^n}\cdot \frac{n + 6^{n}}{n + 1 + 6^{n+1}} $$ $$ = \lim_{n \to +\infty}\frac{\frac{n+1}{4^n}+ 4}{\frac{n}{4^n} + 1}\cdot \frac{\frac{n}{6^n}+ 1}{\frac{n+1}{6^n} + 6} = \frac{2}{3} < 1 $$ The series converge.

Mathsource
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It is good to get good knowledge of the size of the terms involved.

For instance, you may have noticed that $4^n$ is way bigger than $n$.

You can elaborate as follows:

$$n+4^n=4^n(\frac{n}{4^n}+1),$$ and note that the second factor tends to $1$. (This is what you do with rational functions of $n$,when you calculate limits).

If you proceed likewise with the denominator, you have the same thing. Thus, if you call your initial series $a_n$, you have that

$a_n\sim (4/6)^n$, where this means that $lim\ \frac{a_n}{(2/3)^n}=1$.

Even if parameters would change, this solution will essentially work out.

Theon Alexander
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