1. Given $\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)}$ does it converge or diverge?
  2. Given $\sum_{n=1}^{\infty}\frac{5^{n}+4^{n}}{7^{n}-2^{n}}$ does it converge or diverge?

For 1. I know that $\sum_{n=1}^{\infty}\frac{1}{n(n+1)} \rightarrow 1$. Therefore the first series converges by comparison test (since in 1 the terms are smaller.)

For 2. It seems a general strategy for exponential series is to try and compare it to a geometric series. Dividing by the highest exponential I get $\sum_{n=1}^{\infty}\frac{(5/7)^{n} + (4/7)^{n}}{1 - (2/7)^{n}}$. I get stuck now since the denominator is being made smaller, I think this might diverge, but I'm not sure what to compare it too.

Thanks all. A few different methods for solving this. I'll also take a look at asymptotic comparison test.

  • $\begingroup$ $\frac{5^{n}+4^{n}}{7^{n}-2^{n}}\sim\bigl(\frac 57\bigr)^n$ $\endgroup$ – Fabio Lucchini Nov 16 at 9:50
  • $\begingroup$ Why can we ignore the $4^{n}$ and $2^{n}$ is it because the $5^{n}$ and $7^{n}$ are the highest powers? $\endgroup$ – AnalysisLearner Nov 16 at 9:55
  • $\begingroup$ $5^n$ is substantially larger than $4^n$. You can use binomial expansion to see that as $n\to\infty$, $5^n$= $4^n +$ infinitely other very large terms. $\endgroup$ – Certainly not a dog Nov 16 at 10:21

For the second:

  • $5^{n}+4^{n} < 2 \cdot 5^n$
  • $7^{n}-2^{n} > \frac12 \cdot 7^n$
  • $\dfrac{5^{n}+4^{n}}{7^{n}-2^{n}} < 4\cdot \left(\dfrac57\right)^n$
  • $\sum\limits_{n=1}^{\infty}\dfrac{5^{n}+4^{n}}{7^{n}-2^{n}} < 4\cdot \sum\limits_{n=1}^{\infty}\left(\dfrac57\right)^n = 4 \cdot \dfrac52 =10$

so it converges, in fact to just under $4.44$


Note that as $n\to\infty$ we have \begin{align} &n(n+1)(n+2)\sim n^3& &5^n+4^n\sim 5^n& &7^n-2^n\sim 7^n \end{align} hence \begin{align} &\frac 1{n(n+1)(n+2)}\sim\frac 1{n^3}& &\frac{5^n+4^n}{7^n-2^n}\sim\left(\frac 57\right)^n \end{align} hence asymptotic comparison test applies.

  • $\begingroup$ Thanks, what is asymptotic comparison test? I've only learnt the direct comparison test and the p-test. $\endgroup$ – AnalysisLearner Nov 16 at 10:00
  • $\begingroup$ See here. $\endgroup$ – Fabio Lucchini Nov 16 at 10:03
  • $\begingroup$ @AnalysisLearner: This is the equivalence test (comes from asymptotic analysis). $\endgroup$ – Bernard Nov 16 at 10:36
  • $\begingroup$ @Bernard I can't seem to find this test when I google, and the link Fabio provided above seems to only show the comparison test that I already know? $\endgroup$ – AnalysisLearner Nov 16 at 10:38
  • $\begingroup$ It's only a corollary of the general comparison test. The same is true if $ \lim_n \frac{a_n}{b_n}=q$, $0<q<\infty$: $(a_n)$ and $(b_n)$ both converge or both diverge. The same is true for series with (ultimately) terms of constant sign. $\endgroup$ – Bernard Nov 16 at 10:50

For the second series, notice that:

$$\frac{5^{n}+4^{n}}{7^{n}-2^{n}} = \frac{5^{n}\left(1+\left(\frac{4}{5}\right)^{n}\right)}{7^{n}\left(1-\left(\frac{2}{7}\right)^{n}\right)}.$$

It is clear that:

  • $1+\left(\frac{4}{5}\right)^{n} \leq 1 + \frac{4}{5} = \frac{9}{5} ~\forall n \in \mathbb{N}.$
  • $1-\left(\frac{2}{7}\right)^{n} \geq 1 - \frac{2}{7} = \frac{5}{7} ~\forall n \in \mathbb{N}.$

As a consequence:

$$\frac{5^{n}\left(1+\left(\frac{4}{5}\right)^{n}\right)}{7^{n}\left(1-\left(\frac{2}{7}\right)^{n}\right)} \leq \frac{5^{n} \cdot \frac{9}{5}}{7^{n} \cdot \frac{5}{7}} = \frac{63}{25} \cdot \left(\frac{5}{7}\right)^n \sim \left(\frac{5}{7}\right)^n.$$


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