Convergence of series using ratio test.

For the series $$\sum_{n=1}^{\infty}{\frac{n^2}{(n+1)(n^2+2)}},$$ application of the ratio test gave me a result of the limit equal to $$1.$$ This says that the test is inconclusive and should test for convergence using other ways. So I used the limit comparison test next. What I am not understanding is how did the answer in the book arrive at using $$\sum\frac 1n$$ as the comparison series. Can someone explain this for a calc 2 student. Please and thank you.

• – Shaun
Commented Apr 23, 2023 at 16:05
• Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.
– Community Bot
Commented Apr 23, 2023 at 16:55
• How did they arrive at $1/n$? They noted that $\frac{n^2}{(n+1)(n^2+2)} \sim \frac{1}{n}$ as $n \to \infty$. Commented Apr 23, 2023 at 17:02
• I'm a bit confused by "So I used the limit comparison test next," since that needs something to compare it to. Do you mean "I wanted to use the limit comparison test next"? Otherwise, what was the other sequence you used for the limit comparison test? Commented Apr 24, 2023 at 0:58

When dealing with convergence of series, what really matters is the end behaviour of the series.

In this case, the $$n$$th term of the series is the rational function

$$f(n)=\frac{n^2}{(n+1)(n^2+2)}.$$

Therefore, as $$n\to\infty,$$ we see that $$n+1$$ becomes arbitrarily close to $$n$$ and $$n^2+2$$ becomes arbitrarily close to $$n^2,$$ so that $$f(n)$$ becomes arbitrarily close to $$\frac{n^2}{n\cdot n^2}=\frac 1n.$$ This ensures that the given series behaves like the harmonic series towards the tail.

This is called asymptotic comparison of functions, if you want to look further. But that's just the basic idea, as you asked.

I've developed an approach that I think my students understood (well, the ones that come see me). I relate it back to calculus I where we first encounter limits, and the idea is that of "highest order terms."

When you compute limits at infinity, we usually show students to look for the "fastest" growing function. The basic hierarchy is $$\ln x < x^n < b^x < x!$$

but we now focus on polynomials. Consider our limit

$$L = \lim_{n \to \infty} \frac{n^2}{(n+1)(n^2 + 2)}$$

and notice that we can expand the denominator and write

$$L = \lim_{n \to \infty} \frac{n^2}{n^3+n^2 + 2n + 2}.$$

Now observe that the highest power that appears is $$n^3$$, so we divide both numerator and denominator by $$n^3.$$ Thus,

$$L = \lim_{n \to \infty} \frac{\frac{1}{n}}{1 + \frac{1}{n} + \frac{2}{n^2} + \frac{2}{n^3}}$$

and notice that all terms $$\frac{k}{n^p} \to 0$$ as $$n \to \infty.$$ The function behaves similar to $$\frac{1}{n}$$, and this gives us a clue to use LCT against it.

You can even be a bit more hand-wavy and observe that if $$n$$ is large, it's no different than $$n+1$$ and in a similar vain, $$n^2 + 2$$ is no different than $$n^2.$$ You can then "discard" the constants and only leave the highest powers and arrive at the same guess.

• When you say all terms k/n^p --> 0 as n -->∞. So this would lead to the entire expression becoming something like 0/1 wouldn't it? And because of this you are saying that in the infinite sense, this function behaves like 1/n, correct? Thanks Commented Apr 24, 2023 at 23:00
• In a sense, yes, if I understand you correct. I avoided using the notation $f \sim g$ which we read as "The function $f$ is asymptotic to $g$," which means $\frac{f(x)}{g(x)} \to K$ as $x \to \infty,$ for some constant $K.$ Commented Apr 25, 2023 at 0:26

At first we can observe that

• numerator is equal to $$n^2$$
• denominator is equal to $$(n+1)(n^2+2)=n^3+n^2+2n+2$$

and in the latter for $$n$$ "large" the $$n^3$$ term dominates, therefore we have that as $$n\to \infty$$

$$\frac{n^2}{(n+1)(n^2+2)}=\frac{n^2}{n^3+n^2+2n+2}\sim \frac {n^2}{n^3}=\frac1n$$

which indicates that the given series diverges.

More formally one way to proceed is by limit comparison test that is

$$\frac{\frac{n^2}{(n+1)(n^2+2)}}{\frac1n}=\frac{n^3}{(n+1)(n^2+2)}\to 1$$

As an alternative, another way is by direct comparison test that is eventually

$$\frac{n^2}{(n+1)(n^2+2)} \ge \frac{n^2}{(n+n)(n^2+n^2)}= \frac{n^2}{2n\cdot 2n^2}=\frac14 \frac1n$$

and in both case we can conclude that the given series behaves like $$\sum\frac 1n$$ which diverges.

• @Teepeemm Oh yes, I've read again the question and it seems in this way. I revise it, thanks!
– user
Commented Apr 24, 2023 at 7:03