Showing that the series $\sum\left(\frac{n+(-1)^n}{2n+10}\right)^\frac{n}{3}$ is divergent/convergent

Question

I have to decide to if this series $$\sum_{n=0}^\infty\left(\frac{n+(-1)^n}{2n+10}\right)^\frac{n}{3}$$ is divergent or convergent. Would be really nice if someone could help me, as I feel a little bit lost, and so that I can do the other tasks alone, hopefully.

Answer 1

First of all, I would always advice to check whether the necessary condition for convergence is satisfied: if a series converges, then the general term of the sum vanishes as $n$ approaches infinity. In this particular case, this condition is satisfied, so our series could converge.

Now it's time to really check the convergence. As the above comment suggests, we have a lot of tools, in this particular case I would choose the root test (the reason will be soon clear):

Root test

Let $\sum_{n} a_n$ be a series. Let $$ l= \limsup_{n \rightarrow +\infty} |a_n|^{1/n} $$

Then:

(i) If $l < 1$, the series converges absolutely.

(ii) If $ l> 1$, the series diverges.

(iii) If $l=1$, the test in inconclusive (i.e, the series could converge or diverge, we need further inspections).

If you're familiar with limits, you should be able to conclude now. Note that the root test is not always necessarily the best choice, but in this case the form of $a_n$ really suggests it.

Answer 2

Because I could use the practice, using the root test as suggested above, we seek to evaluate the limit $$ \lim_{n\rightarrow \infty}|\frac{n+(-1)^n}{2n+10}|^{1/3} $$ let's bound $$ |\frac{n+(-1)^n}{2n+10}|^{1/3}\leq |\frac{n+(-1)^n}{2n}|^{1/3}\stackrel{\text{triangle ineq.}}{\leq}[\frac{n+|-1|}{2n}]^{1/3}\\ =(\frac{1}{2}+\frac{1}{2n})^{1/3} \rightarrow \frac{1}{2^{1/3}}<1 $$ as $n\rightarrow \infty$, and so the series converges.

Answer 3

Me, I generally try to avoid the "mechanical" tests whenever possible. Here I notice that

$$\tag 1 \frac{n+(-1)^n}{2n+10} \to \frac{1}{2}\, \text { as } n\to \infty.$$

Thus the terms in $(1)$ are between $0$ and $2/3$ for large $n.$ This tells us the terms of our series are positive and bounded above by $(2/3)^{n/3}=[(2/3)^{1/3}]^n$ for large $n.$ Because $(2/3)^{1/3}<1,$ $\sum_n [(2/3)^{1/3}]^n$ is a convergent geometric series. Our series therefore converges by the comparison test.