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I had a standard problem in my textbook which was to determine the convergence of $\sum _{n=2}^\infty\frac{n^3+1}{n^4-1}$. To determine whether the series is convergent or not the standard solution seems to be the comparison test:

$$\sum _{n=2}^\infty\frac{n^3}{n^4} \le \sum _{n=2}^\infty\frac{n^3+1}{n^4-1}$$

The series on the left is equal to $\sum _{n=2}^\infty\frac{1}{n}$ which divergence I may asssume due to the so-called p-series test. It follows that the series on the right must also be divergent by the comparison test.

However, I started wondering what if I had to determine convergence/divergence of a similar series:

$$\sum _{n=2}^\infty\frac{n^3-1}{n^4+1}$$

Notice the switched plus and minus signs. Now I intuitively see that the series is supposed to diverge as well but I can't apply the comparison test as I did before. What then is the easiest way to prove it? I tried the ratio test but the result is 1 which is inconclusive.

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4 Answers 4

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Either you can use the Limit Comparison Test and say that

$$\lim_{n \to \infty} \frac{\frac 1n}{\frac{n^3-1}{n^4+1}} =1 $$

So, $\sum \frac 1n$ and $\sum \frac{n^3-1}{n^4+1}$ both convergent or both divergent.

Or you can use the direct comparison test:

$$\frac{n^3-1}{n^4+1} \geq \frac{\frac 12 n^3}{n^4 + n^4} = \frac{n^3}{4n^4} = \frac{1}{4n}$$

Using the p-test, the result follows.

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  • $\begingroup$ (+1) The limit comparison test is a much better answer here than coming up with some ad-hoc comparison test. The same method can show the convergence or divergence of any series whose general term is a rational function of $n$. $\endgroup$
    – Jack M
    Commented May 14, 2016 at 21:59
  • $\begingroup$ Since the OP does not seem familiar with the limit comparison test, though, stating its assumptions may be useful. (In particular, that the sequences must have constant sign for $n$ big enough). $\endgroup$
    – Clement C.
    Commented May 14, 2016 at 21:59
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The standard way is to compare with the series $\sum 1/n$ using limits: $$ \lim_{n\to\infty}\frac{\dfrac{n^3-1}{n^4+1}}{\dfrac1n}=1. $$ since both series are of positive terms and the limit exists and is $\ne0,\infty$, both series are of the same character, convergent or divergent,

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You may compare it to $$ \sum_{n=2}^\infty \frac{n^3}{4n^4} $$ since $\frac{n^3 -1}{n^4+1} \geq \frac{n^3 - \frac{n^3}{2}}{n^4+n^4}$.


More generally, a very useful strengthening of the comparison test: if $(a_n)_n, (b_n)_n$ are two positive sequences such that $\frac{a_n}{b_n} \xrightarrow[n\to\infty]{} 1$, then $\sum_n a_n$ converges iff $\sum_n b_n$ does.

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  • $\begingroup$ Note that assumption of the theorem stated at the end that $a_n \geq 0$ for $n$ big enough is necessary. Indeed, a counterexample would be $a_n = \frac{(-1)^n}{\ln n}$ and $b_n = a_n+ \frac{1}{n}$. $\endgroup$
    – Clement C.
    Commented May 14, 2016 at 22:02
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For example: since $\;2n^7-2n^4\ge n^7+n^3\;$ (Can you prove this? It is true for $\;n\ge2\;$) , we get

$$\frac{n^3-1}{n^4+1}\ge\frac{n^3}{2n^4}=\frac1{2n}$$

so again by comparison out series diverges.

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