Direct and limit comparison for positive series

Understand the idea

Comparison tests use a series whose behavior is already known to study another positive series. Direct comparison needs an inequality in a useful direction. Limit comparison instead shows that two positive term formulas have the same eventual scale.

To prove convergence by direct comparison, bound the target above by a convergent positive series. To prove divergence, bound it below by a divergent one. The opposite inequalities do not give those conclusions. Conditions may hold only after a finite starting index.

For limit comparison, choose a benchmark that matches the dominant powers and calculate the ratio explicitly. A finite positive ratio makes the two series share convergence behavior. Limits of zero or infinity require a separate argument rather than the same automatic conclusion.

0<a_n\le b_n,\ \sum b_n<\infty\Rightarrow\sum a_n<\infty;\qquad \lim\frac{a_n}{b_n}=L\in(0,\infty)\Rightarrow\text{same behavior}

A method you can apply

Choose a positive benchmark, often a p-series or geometric series.
Establish a useful inequality or compute a finite positive limit of term ratios.
State the benchmark's behavior and the precise conclusion transferred to the target.

Worked example

Determine the convergence of $\displaystyle \sum_{n=1}^{\infty}(3n+2)/(n^2+1)$ .

The dominant powers suggest comparing with the harmonic terms $\displaystyle b_n=1/n$ .
The ratio is $\displaystyle n(3n+2)/(n^2+1)$ , which approaches $\displaystyle 3$ .
The limit is finite and positive, and the harmonic series diverges.

The target series diverges by limit comparison with

\displaystyle \sum1/n

.

Try it yourself

Determine the convergence of $\displaystyle \sum_{n=1}^{\infty}5/(n^2+7)$ .

Show solution and reasoning

For $\displaystyle n\ge1$ , $\displaystyle 0<5/(n^2+7)\le5/n^2$ . The upper comparison series is a constant multiple of a convergent p-series with $\displaystyle p=2$ , so the target converges.