Convergence of Infinite Series

Adding $1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots$ gives partial sums $1$, $1.5$, $1.75$, $1.875$, visibly closing on 2, and the limit is exactly 2. Adding $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots$ gives partial sums that pass every bound eventually, though slowly: a hundred thousand terms reach only about $12.09$.

The terms shrink to zero in both cases, and they differ in how fast. That rate decides whether the sum converges.

An infinite sum is not an addition performed forever. It is the limit of the partial sums, defined exactly as the improper integral was, notation for a limit, which may or may not exist. The convergence tests are the tools for deciding, and none of them is a single universal criterion, which is why choosing the right one is part of the skill.

The canonical definition lists the tests. What follows is why each is stated with its particular hypotheses, and what it leaves undecided.

Why the $n$th-term test proves only divergence. If $\sum a_k$ converges to $s$, then $a_n=s_n-s_{n-1}\to s-s=0$. So convergence forces the terms to vanish, and the contrapositive, terms not vanishing implies divergence, is the test. The converse is false, and the harmonic series is why: its terms vanish and its partial sums grow like $\ln n$, reaching $12.09$ after ${10}^5$ terms and passing every bound eventually.

A test that can only ever say "diverges" is worth applying first precisely because it is cheap.

Why the ratio test is inconclusive at $L=1$. The test works by comparison with a geometric series of ratio $L$. When the terms are eventually dominated by a convergent geometric series; when they eventually grow. At $L=1$ the comparison supplies nothing, and both outcomes genuinely occur: every $p$-series has $lim|a_{n+1}/a_n|=lim{\left(\frac{n}{n+1}\right)}^p=1$, yet $\sum n^{-2}$ converges and $\sum n^{-1}$ does not.

So $L=1$ is not a gap in the proof. It is a region where the ratio carries no information, and a different test is required.

Why the integral test needs positive and decreasing terms. The proof brackets the partial sums between ${\int }_1^{n+1}f$ and $a_1+{\int }_1^{n}f$, which requires the rectangles to sit consistently above or below the curve. Without monotonicity the comparison collapses. With it, the series and the improper integral converge together, which is why the $p$-series threshold is the same number as the improper-integral threshold from the previous unit, read on the same function.

Why absolute convergence is a stronger claim. If $\sum |a_n|$ converges then so does $\sum a_n$, but not conversely. The alternating harmonic series converges to $\ln 2\approx 0.693147$, its partial sums reach $0.69314468$ after $2\times {10}^5$ terms, while the harmonic series of its magnitudes diverges. The cancellation between positive and negative terms is doing the work.

That dependence on cancellation is fragile: a conditionally convergent series can be rearranged to sum to any real number at all, so its "sum" depends on the order of addition. Absolute convergence removes that dependence, which is why it is the condition under which the usual algebraic manipulations are safe.

The geometric series. Write the partial sum and multiply by $r$:

Subtracting, every interior term cancels:

This is exact for every $n$. No limit has been taken yet. Now let $n\to \mathrm{\infty }$. The only $n$-dependent piece is $r^{n+1}$, which tends to 0 exactly when . So

and for $|r|\ge 1$ the term $r^{n+1}$ does not vanish and the series diverges.

Verification. At $r=0.5$ the sixty-term partial sum is $2.0000000000$ against $\frac{1}{0.5}=2$; at $r=0.9$ it is $9.98203$ against 10, still climbing; at $r=-0.5$ it is $0.6666666667$ against $\frac{1}{1.5}$.

The condition is not a technical restriction. It is precisely the condition under which the limit exists.

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.

---

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.

---

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.

1. A geometric series: ${\displaystyle \sum _{k=0}^{\mathrm{\infty }}\frac{3}{4^k}}$.

Write it as $3\sum (1/4{)}^k$, so $a=3$ and $r=\frac{1}{4}$. Since it converges, to

Check: partial sums $3$, $3.75$, $3.9375$, $3.984375$ climb toward 4.

2. The $n$th-term test: ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{n}{2n+1}}$.

The terms tend to $\frac{1}{2}$, not 0. A convergent series must have vanishing terms, so this one diverges, decided in one line without any further test.

3. Terms vanish and the series still diverges: ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}}$.

The $n$th-term test is silent here, since $1/n\to 0$. Use the integral test: $f(x)=1/x$ is positive and decreasing on $[1,\mathrm{\infty })$, and

from the previous unit. So the series diverges too.

How slowly: the partial sum after ${10}^5$ terms is about $12.090$, against $\ln ({10}^5)=11.513$. The excess is the Euler–Mascheroni constant, and the growth is unbounded but logarithmic.

4. The ratio test: ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{2^n}{n!}}$.

Since , the series converges absolutely. The factorial beats the exponential, which is the same comparison L'Hôpital established for $x^n/e^x$.

5. Conditional convergence: ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n+1}}{n}}$.

The magnitudes $1/n$ decrease to 0, so by the alternating series test it converges. But the series of magnitudes is the harmonic series, which diverges, so the convergence is conditional, resting on cancellation.

The value: the partial sum after $2\times {10}^5$ terms is $0.69314468$, against $\ln 2=0.69314718$. Convergence is slow precisely because the cancellation is doing all the work.

Step 1 — check the terms. Compute $lim{a}_n$. If it is not 0, the series diverges and no further work is needed. If it is 0, the test is silent and you must continue. This step can never establish convergence.

Step 2 — recognise a standard form.

Most exercises are one of these or reduce to one.

Step 3 — for factorials or $n$th powers, use the ratio test. The signature is $n!$, $a^n$, or $n^n$ in the term. Compute $L=lim|a_{n+1}/a_n|$: below 1 converges absolutely, above 1 diverges, and exactly 1 decides nothing, move to another test rather than guessing.

Step 4 — for a term resembling a function you can integrate, use the integral test. Confirm the terms are positive and decreasing first; both are hypotheses, not formalities. The series and ${\int }_1^{\mathrm{\infty }}f$ then share a verdict.

Step 5 — otherwise compare. Bound the term above by something known to converge, or below by something known to diverge. The $p$-series and the geometric series are the usual standards.

Step 6 — for an alternating series, ask which kind of convergence. Test $\sum |a_n|$ as well. Convergence of both means absolute; convergence of the alternating series alone means conditional, and a conditionally convergent series may not be rearranged.

Checks. Compute a few partial sums and see whether they settle. A series whose terms do not vanish needs no further test; one whose terms do vanish needs one.

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.