Convergence of Infinite Series

Each series below is decided by a different test. The point is the selection, not the arithmetic: reading the shape of a term tells you which test can settle it.

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1. ${\displaystyle \sum _{k=0}^{\mathrm{\infty }}\frac{3}{4^k}}$ — geometric, so sum it. Ratio $r=\frac{1}{4}$ with , so it converges to $\frac{3}{1-1/4}=4$. Check: partial sums $3$, $3.75$, $3.9375$. This is the only shape whose sum comes in closed form, which is why it is worth recognising first.

2. ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{n}{2n+1}}$ — the $n$th-term test, one line. The terms tend to $\frac{1}{2}$, not $0$. A convergent series must have vanishing terms, so this diverges. No further work. Always try this first: it is the cheapest test and it can only ever say "diverges".

3. ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}}$ — terms vanish and it still diverges. The $n$th-term test is silent. The integral test settles it: $f(x)=1/x$ is positive and decreasing on $[1,\mathrm{\infty })$, and ${\int }_1^{\mathrm{\infty }}dx/x$ diverges, so the series does too. How slowly: the partial sum after ${10}^5$ terms is about $12.090$, against $\ln ({10}^5)=11.513$. Unbounded, but logarithmically.

4. ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{2^n}{n!}}$ — a factorial, so the ratio test.

, so it converges absolutely. The signature to look for is $n!$, $a^n$ or $n^n$ in the term.

5. ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2+3n}}$ — comparison. For $n\ge 1$, $\frac{1}{n^2+3n}\le \frac{1}{n^2}$, and $\sum n^{-2}$ is a convergent $p$-series with . Bounding above by a convergent series establishes convergence. Note the direction: bounding above by a divergent series, or below by a convergent one, establishes nothing at all.

6. ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n+1}}{n}}$ — alternating, so ask which kind of convergence. The magnitudes $1/n$ decrease to $0$, so by the alternating series test it converges, to $\ln 2\approx 0.693147$. But the magnitudes form the harmonic series of case 3, which diverges. So the convergence is conditional, resting entirely on cancellation. The value: after $2\times {10}^5$ terms the partial sum is, converging slowly because the cancellation is doing all the work.

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Reading the table backwards.

Cases 3 and 6 use the same terms in magnitude and reach opposite verdicts, which is the whole content of the absolute-conditional distinction. Cases 4 and 5 both converge and could not be settled by each other's test: the ratio test on case 5 gives $L=1$ and decides nothing.

Terms tending to zero does not give convergence. The harmonic series $\sum 1/n$ has $a_n\to 0$ and diverges. Its partial sums grow like $\ln n$: about $12.09$ after ${10}^5$ terms, against $\ln ({10}^5)=11.51$. The growth is slow enough to look like settling and unbounded all the same.

The $n$th-term test runs one way only, non-vanishing terms prove divergence, vanishing terms prove nothing.

The ratio test at $L=1$ decides nothing. Both $\sum 1/n$ and $\sum 1/n^2$ give $L=lim{\left(\frac{n}{n+1}\right)}^p=1$, and the first diverges while the second converges. Reporting "the ratio test gives 1, so the series diverges" asserts something the test never said. A different test is required, and for these the integral test settles both.

A convergent series cannot always be rearranged. The alternating harmonic series converges to $\ln 2\approx 0.693147$. Because the convergence is conditional, the magnitudes form the divergent harmonic series, its terms can be reordered to sum to any real number, or to diverge. So "the sum" depends on the order of addition, and only absolute convergence removes that dependence.

If the terms of a series tend to zero then the series converges, so $\sum 1/n$ converges because $1/n\to 0$.