How to Show \(\sum \frac{n-1}{n^3+1}\) Converges

Expert intro: This is a classic comparison problem. The numerator grows like $n$, while the denominator grows like $n^3$, so the term behaves like $1/n^2$.

Detailed walkthrough

Step 1: Identify a comparison series

Let

$a_n=\frac{n-1}{n^3+1}.$

For large $n$, this behaves like

$\frac{n}{n^3}=\frac{1}{n^2}.$

So a natural comparison is

$b_n=\frac{1}{n^2}.$

Step 2: Check the inequality

For $n\ge 1$, we can compare directly:

$\frac{n-1}{n^3+1}\le \frac{n}{n^3}=\frac{1}{n^2}.$

The key reason is that $n^3+1\ge n^3$ and $n-1\le n$.

Step 3: Use the p-series test

We know

$\sum_{n=1}^{\infty}\frac{1}{n^2}$

converges because it is a p-series with $p=2>1$.

Since

$0\le \frac{n-1}{n^3+1}\le \frac{1}{n^2},$

the comparison test gives

$\sum_{n=1}^{\infty}\frac{n-1}{n^3+1} \text{ converges.}$

💡 Pitfall guide

Don't try to compare it with $1/n$ just because the numerator has an $n$ in it. The denominator has degree 3, so the whole fraction is much smaller than $1/n$. The sharper comparison is with $1/n^2$.

🔄 Real-world variant

If the denominator were $n^2+1$ instead of $n^3+1$, then the term would behave like $(n-1)/n^2\sim 1/n$, and the conclusion could change to divergence. If the numerator were $n^2-1$, then the term would behave more like $1/n$ as well, so degree counting matters.

🔍 Related terms

p-series, comparison test, term growth