Series Convergence: Three Decisions to Explain

Interpret the ratio-test limit

The terms are positive, so the absolute-value ratio is the same as $a_{n+1}/a_n$. Substituting the terms and expanding the denominator gives

$$\frac{n^2+3n}{n^2+5n+4}.$$

Divide the numerator and denominator by $n^2$:

$$\frac{1+3/n}{1+5/n+4/n^2}.$$

As $n\to\infty$, the numerator and denominator both approach 1, so the ratio approaches 1.

When this limit is $1$, the ratio test is inconclusive: it proves neither convergence nor divergence. Use another test.

Find a convergent upper comparison

For every $n\geq1$, we have $n^2+3n\geq n^2>0$. Taking reciprocals reverses the inequality, so

$$0<\frac{1}{n^2+3n}\leq\frac{1}{n^2}.$$

A p-series has terms $1/n^p$, where $p$ is a fixed real exponent; it converges when $p>1$. Here $p=2$, so the comparison series converges.

State what the comparison proves

The original terms are nonnegative and no larger than the corresponding terms of a convergent series. Direct comparison therefore proves that the original series converges. Because all its terms are positive, taking absolute values leaves the series unchanged: it also converges absolutely.

The decisive evidence is the convergent upper comparison, not the inconclusive ratio-test limit.

Explain why the upper bound is insufficient

The upper bound is correct, but it does not establish divergence. The terms of a convergent series can also be smaller than those of a divergent series: for example, $1/n^2\leq1/n$. To prove divergence by direct comparison here, find a lower bound supplied by the terms of a divergent positive series.

Build a divergent lower comparison

Since $n\geq1$, we have $4\leq4n^2$, so $n^2+4\leq5n^2$. The denominators are positive; taking reciprocals reverses the inequality. Multiplying by the positive number $n$ gives

$$b_n=\frac{n}{n^2+4}\geq\frac{1}{5n}.$$

Compare the partial sums

The harmonic series, $1+1/2+1/3+\cdots$, has partial sums that grow without bound. The partial sums of the series with terms $1/(5n)$ are one fifth as large and also grow without bound. The corresponding partial sums of the original series are at least as large.

The original series therefore diverges to $+\infty$. A divergent lower comparison forces this conclusion; the student's divergent upper comparison did not.

Check the alternating-series conditions

The series begins $1/3-1/4+1/5-\cdots$. Its signs alternate. Its magnitudes are positive and decrease because, for every $n\geq1$,

$$0<\frac{1}{n+3}<\frac{1}{n+2}.$$

They also tend to zero:

$$\lim_{n\to\infty}\frac{1}{n+2}=0.$$

These conditions establish convergence by the alternating-series test.

Test the series of absolute values

Absolute convergence means that the series formed by taking the absolute value of every term converges. Here its terms are $|c_n|=1/(n+2)$. For $n\geq1$, we have $n+2\leq3n$, and both denominators are positive. Therefore

$$\frac{1}{n+2}\geq\frac{1}{3n}.$$

The series with terms $1/(3n)$ diverges because its partial sums are one third of the harmonic partial sums. Direct comparison shows that the series of absolute values also diverges.

Combine the two conclusions

A series is conditionally convergent when it converges but its series of absolute values diverges. The original alternating series is therefore conditionally convergent.

Absolute convergence would imply convergence. Here the alternating-series test established convergence, but it did not establish absolute convergence; the comparison above rules that out.