Type a command or search the notes. Tab cycles completions. Up and down recall history when blank. Escape dismisses, then leaves the prompt.
MATH
[1251]

Theorems from Tests for series convergence

All theorems from Tests for series convergence.

Back to Tests for series convergence

Integral test

Suppose ff is positive, continuous and decreasing for x≥Nx\ge N, where NN is a positive integer and an=f(n)a_n=f(n). Then

∑n=N∞an\sum_{n=N}^{\infty}a_n

and

∫N∞f(x) dx\int_N^\infty f(x)\,dx

either both converge or both diverge.

Positivity and monotonicity give, for m≥Nm\ge N,

∫Nm+1f(x) dx≤∑n=Nmf(n)≤f(N)+∫Nmf(x) dx.\int_N^{m+1}f(x)\,dx \le\sum_{n=N}^{m}f(n) \le f(N)+\int_N^{m}f(x)\,dx.

The integral and partial sums therefore have the same convergence behaviour.

  1. [1]:Infinite series
  2. [2]:The pp-series test
  3. [3]:A logarithmic borderline

The pp-series test

For real pp,

∑n=1∞1np{converges,p>1,diverges,p≤1.\sum_{n=1}^{\infty}\frac1{n^p} \begin{cases} \text{converges},&p>1,\\ \text{diverges},&p\le1. \end{cases}

When p≤0p\le0, the terms do not tend to zero. When p>0p>0, the integral test uses

∫1∞dxxp,\int_1^\infty\frac{dx}{x^p},

which is finite exactly when p>1p>1. The special case p=2p=2 is evaluated exactly in Euler's Basel sum.

  1. [1]:Integral test

Direct comparison test

Suppose 0≤an≤bn0\le a_n\le b_n for all sufficiently large nn. Then:

  • if ∑bn\sum b_n converges, ∑an\sum a_n converges;
  • if ∑an\sum a_n diverges, ∑bn\sum b_n diverges.

The inequality must point towards the conclusion. A smaller positive series is controlled by a convergent larger series; a larger positive series inherits divergence from a divergent smaller series.

  1. [1]:Infinite series
  2. [2]:Direct comparisons

Limit comparison test

If an>0a_n>0 and bn>0b_n>0 for all sufficiently large nn and

lim⁡n→∞anbn=K\lim_{n\to\infty}\frac{a_n}{b_n}=K

with 0<K<∞0<K<\infty, then

∑an converges⟺∑bn converges.\sum a_n\text{ converges} \quad\Longleftrightarrow\quad \sum b_n\text{ converges}.

The finite positive ratio means that ana_n and bnb_n are eventually bounded above and below by positive multiples of one another. If the ratio is 00, only one-way conclusions are available: convergence of ∑bn\sum b_n implies convergence of ∑an\sum a_n and divergence of ∑an\sum a_n implies divergence of ∑bn\sum b_n.

  1. [1]:Infinite series
  2. [2]:Dominant-term comparisons
  3. [3]:Sequences and series

Ratio test

For a series of positive terms, suppose

lim⁡n→∞an+1an=L.\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=L.

Then:

  • L<1L<1 implies convergence;
  • L>1L>1 implies divergence;
  • L=1L=1 gives no conclusion.

For sign-changing terms, apply the test to ∣an∣|a_n| when testing absolute convergence.

  1. [1]:Infinite series
  2. [2]:Factorials and exponentials

Root test

If

lim⁡n→∞∣an∣n=L,\lim_{n\to\infty}\sqrt[n]{|a_n|}=L,

then:

  • L<1L<1 implies absolute convergence;
  • L>1L>1 implies divergence;
  • L=1L=1 gives no conclusion.

If the ratio test gives 11, switching to the root test does not automatically help: the root test may also be inconclusive.

  1. [1]:Infinite series
  2. [2]:Large powers
  3. [3]:Sequences and series
  4. [4]:A power series and its endpoints
  5. [5]:Seen Questions

Absolute convergence implies convergence

If ∑∣an∣\sum|a_n| converges, then ∑an\sum a_n converges.

If SnS_n are the partial sums, then for m>nm>n

∣Sm−Sn∣=∣∑k=n+1mak∣≤∑k=n+1m∣ak∣.|S_m-S_n| =\left|\sum_{k=n+1}^{m}a_k\right| \le\sum_{k=n+1}^{m}|a_k|.

The tails of the convergent series ∑∣ak∣\sum|a_k| become arbitrarily small, so any two sufficiently late partial sums of ∑ak\sum a_k are arbitrarily close. Completeness of R\mathbb R then guarantees that these partial sums approach a finite limit, even when the signs of the terms vary.

  1. [1]:Absolute and conditional convergence

Alternating-series test

If bn≥0b_n\ge0, bn+1≤bnb_{n+1}\le b_n and bn→0b_n\to0, then

∑n=0∞(−1)nbn\sum_{n=0}^{\infty}(-1)^n b_n

converges.

The successive partial sums lie on opposite sides of the limiting value and close in on it. The magnitudes need only be eventually decreasing, since finitely many initial terms do not affect convergence.

  1. [1]:Infinite series
  2. [2]:Conditional convergence of the alternating harmonic series
  3. [3]:A power series and its endpoints
  4. [4]:Seen Questions

Rearranging a series

Every rearrangement of an absolutely convergent series has the same sum. A conditionally convergent series can be rearranged to converge to any prescribed real number, or to diverge to +∞+\infty or −∞-\infty.

The second statement is the Riemann rearrangement theorem.

  1. [1]:Absolute and conditional convergence
  2. [2]:Rearranging the alternating harmonic series

On this page

Integral testThe pp-series testDirect comparison testLimit comparison testRatio testRoot testAbsolute convergence implies convergenceAlternating-series testRearranging a series