Proof that $\frac{n^6+n^2}{n^7+3}$ converges

Question

I'm looking at a problem that follows:

Test if $$\sum_{n=1}^\infty \frac{n^6+n^2}{n^7+3}$$ converges or diverges. I think I have a proof but it seems a bit awkward. You take out the 3 in the denominator and factor the numerator to get $$\sum_{n=1}^\infty \frac{n^6+n^2}{n^7+3} < \sum_{n=1}^\infty \frac{n^2(n^2+1)}{n^7} = \sum_{n=1}^\infty \frac{n^2+1}{n^5}$$ and replace 1 with $n^2$ and thus $$\sum_{n=1}^\infty \frac{n^2+1}{n^5} < \sum_{n=1}^\infty \frac{2n^2}{n^5}$$ which holds for all $n > 1$ and then after cancelling, you can replace 2 with $n$ to get $$\sum_{n=1}^\infty \frac{n}{n^3}$$ which will hold for all $n > 2$ and since that is equal to $\frac{1}{n^2}$, it converges.

Is this proof valid? It seems I've either done too much work to prove this or I am missing some steps.

Answer 1

Use the Limit Comparison Test: compare the given series with $\sum\limits_{n=1}^\infty \frac{1}{n}$.

Answer 2

No. You have shown that a series diverges that is larger than another series that is larger than your series. You could say an upper bound of your series diverges, but that doesn't mean your series diverges. For example:

$$ \sum \frac{1}{n^2} < \sum\frac{1}{n}$$ So does this that $\frac{1}{n^2}$ diverges? To show divergence, we want to create a lower bound for the series and show that it diverges.

$$\sum \frac{n^6+ n^2}{n^7 + 3} > \sum \frac{n^6}{n^7 + 3}$$ Is an example, and then you show that this lower bound diverges through other comparisons. You may do it however you choose. Let me know if you would like more help.

Answer 3

No - recall the (direct) comparison test: suppose $a_n, b_n$ are sequences of positive terms and $a_n \leq b_n$ eventually (that is, there exists $N$ such that $a_n \leq b_n$ for each $n \geq N$). Then if $\sum a_n$ is divergent, so too is $\sum b_n.$

Easier is to use, as @ShyamalSayak suggests, the Limit Comparison Test (I tell my students to do this too). If your argument were to hold, then I can say that $$\sum_{n = 1}^\infty \frac{1}{n^2}$$ is divergent by using the inequality $1 \leq n$ - but it's convergent by the p-series test and the Integral Test!