A Comprehensive Summary of Various Series Tests in Analysis📂Calculus
A Comprehensive Summary of Various Series Tests in Analysis
Overview
This post will introduce several series convergence tests without diving into their proofs. It is often more valuable to utilize these tests as facts, especially since the proofs can be quite tedious.
∃limk→∞xk means that the limit of xk exists in R, meaning it converges. Conversely, ∃limk→∞xk means that the limit of xk does not exist in R, meaning it diverges.
For sufficiently large k, when k→∞limbkak=1 then it is expressed as ak≈bk.
bk↓0 means that bk is a decreasing sequence and converges to 0 while always taking a value greater than or equal to 0.
If limk→∞ak=0 then ∑k=1∞ak diverges:
k→∞limak=0⟹∃k=1∑∞ak
The Divergence Test is the only method you might encounter in high school. It’s a pity that it cannot determine convergence itself, but it’s the easiest and fastest when showing divergence.
Cauchy’s Criterion
The convergence of ∑n=1∞an is equivalent to limn→∞∑k=nn+mak=0:
∃k=1∑∞ak⟺(∀ε>0,∃N∈N:m≥n≥N⟹k=n∑mak<ε)
This is called the Cauchy Criterion because the theorem utilizes the fact that, in the space of real numbers R, a converging sequence being a Cauchy sequence is equivalent.
This section deals with sequences where each term is greater than or equal to 0, denoted ak≥0.
Integral Test
Let’s assume the decreasing functionf:[1,∞)→R is always greater than 0. The convergence of ∑k=1∞f(k) is equivalent to ∫1∞f(x)dx<∞:
∃k=1∑∞f(k)⟺∫1∞f(x)dx<∞
The Integral Test is proven using the fact that f(n+1)≤∫nn+1f(x)dx≤f(n). It’s one of those rare criteria where the proof itself is interesting.
p-Series Test
The convergence of ∑k=1∞k−p is equivalent to p>1:
∃k=1∑∞kp1⟺p>1
The p-Series Test essentially means that if you increase the exponent even slightly in a harmonic series, it converges; otherwise, it diverges. It’s a corollary derived by inserting a geometric series into the integral test, but because it’s so simple and useful, it’s used even more often than the integral test.
Comparison Test
For sufficiently large k, let’s say 0≤ak≤bk. If ∑k=1∞bk converges, then ∑k=1∞ak also converges:
k=1∑∞bk<∞⟹k=1∑∞ak=∞⟹k=1∑∞ak<∞k=1∑∞bk=∞
The Comparison Test compares another series known to converge to determine the convergence as its name suggests. Using the contrapositive, you can similarly check if a series diverges.
Limit Comparison Test
For sufficiently large k, let’s say ak≥0 and bk>0. L:=limk→∞bkak∈R is some extended real number. (1) If 0<L<∞ then both ∑n=1∞an and ∑n=1∞bn either converge or diverge together. (2) If L=0 and ∑n=1∞bn converges, then ∑n=1∞an also converges. (3) If L=∞ and ∑n=1∞bn diverges, then ∑n=1∞an also diverges:
0<L<∞⟹L=0⟹L=∞⟹(∃k=1∑nak⟺∃k=1∑nbk)(∃k=1∑nbk⟹∃k=1∑nak)(∃k=1∑nbk⟹∃k=1∑nak)
The Limit Comparison Test is like the comparison test but for when it’s difficult to show the original series converges by comparing it with another converging series. Though the conditions might seem strenuous, they’re often easy to satisfy when simply proving convergence, making this very useful.
For an infinite series S=∑k=1∞ak, if ∑k=1∞∣ak∣ converges, S is defined to converge absolutely. Accordingly, a series that doesn’t converge absolutely but does converge on its own is said to converge conditionally.
Root Test
For the limit supremumr=limsupk→∞∣ak∣1/k of {∣ak∣1/k}, if r<1 then ∑n=1∞ak absolutely converges, if r>1 then ∑n=1∞ak diverges:
r<1⟹r>1⟹∃k=1∑∞∣ak∣∃k=1∑∞ak
Ratio Test
Assuming ak=0, and r=limk→∞∣ak∣∣ak+1∣∈R is an extended real number. If r<1 then ∑k=1∞ak absolutely converges, if r>1 then ∑k=1∞ak diverges:
r<1⟹r>1⟹∃k=1∑∞∣ak∣∃k=1∑∞ak
The Root Test and Ratio Test have slightly stringent conditions but can outright prove absolute convergence, hence their popularity. Meanwhile, if r=1, methods like Dirichlet’s Test or Alternating Series Test can be used. The alternating series test can be directly derived from Dirichlet’s test, and if you’re only looking to establish convergence, considering an alternating harmonic series as an example is helpful.
Dirichlet’s Test
If the partial sum sn=∑k=1nak is bounded and k→∞ when bk↓0, then ∑k=1∞akbk converges:
∣sn∣<∞,bk↓0⟹∃k=1∑∞akbk
Alternating Series Test
If k→∞ when bk↓0, then ∑n=1∞(−1)kbk converges.
bk↓0⟹∃k=1∑∞(−1)−kbk
Wade. (2013). An Introduction to Analysis(4th Edition): p186, 188. ↩︎
Wade. (2013). An Introduction to Analysis(4th Edition): p193, 194, 196. ↩︎
Wade. (2013). An Introduction to Analysis(4th Edition): p198, 201, 210. ↩︎