# Alternating Series

The factory of the future will have only two employees, a man and a dog. The man will be there to feed the dog. The dog will be there to keep the man from touching the equipment, Warren Bennis

The development of full artificial intelligence could spell the end of the human race, Stephen Hawking.

# Recall

A finite series is given by all the terms of a finite sequence, added together, e.g., {3, 5, 7, . . . , 21}, $\sum_{k=1}^{10} 2k+1 = 120$. An infinite series is the sum of an infinite sequence of numbers. It is represented in the form $\sum_{n=1}^\infty a_n = a_1 + a_2 + a_3 + ···$ where an represents the terms of the sequence, and n is the index that ranges from 1 to infinite.

A series is convergent (or converges) if the sequence of its partial sums tends to a limit, that is, l = $\lim_{n \to ∞} \sum_{k=1}^n a_k$ exists and is a finite number. More precisely, if there exists a number l (or S) such that for every arbitrary small positive number ε, there is a (sufficient large) N, such that ∀n ≥ N, |Sn -l| < ε where Sn = $\sum_{k=1}^n a_k = a_1 + a_2 + ··· + a_n$. If the series is convergent, the number l is called the sum of the series. On the contrary, any series that is not convergent ($\lim_{n \to ∞} \sum_{k=1}^n a_k$ does not exist) is said to be divergent or to diverge.

Divergence Test. If $\lim_{n \to ∞}a_n ≠ 0$, then $\sum_{n=1}^\infty a_n$ diverges.

Integral Comparison. If f(x) is a positive, continuous, and decreasing function on the interval [1, ∞), then $|\sum_{n=1}^\infty f(n) -\int_{1}^{∞} f(x)dx| < f(1).$ Besides, $\sum_{n=1}^\infty f(n)$ converges, if and only if, $\int_{1}^{∞} f(x)dx$ converges.

Theorem. Direct Comparison test. Let {an} and {bn} be positive sequences where an≤bn ∀n≥N, for some N.

1. If $\sum_{n=1}^\infty b_n$ converges, then $\sum_{n=1}^\infty a_n$ converges.
2. If $\sum_{n=1}^\infty a_n$ diverges, then $\sum_{n=1}^\infty b_n$ converges.

Definition. An infinite series is said to be absolutely convergent or to converge absolutely if the absolute values of its terms forms a convergent series, i.e.,  $\sum_{n=0}^\infty |a_n| = L$ for some real number L.

Definition. A series is conditionally convergent if it converges, but does not converge absolutely, e.g., $\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}$

# Alternating Series

Definition. An alternating series is a series whose terms alternate in sign, i.e., $\sum_{n=0}^\infty (-1)^na_n = a_0 -a_1 + a_2 -a_3 + ···$ where either all an are positive or all an are negative.

Theorem. Alternating Series Test. It states that an alternating series converges if the following two conditions are met:

1. |an| decreases monotonically, i.e., |an+1| ≤ |an|.
2. $\lim_{n \to ∞} a_n = 0$.

Other authors express it as follows. Suppose that we have a series ∑an and either an = (-1)nbn or an = (-1)n+1bn, let’s also suppose that

1. bn positive, bn ≥ 0 ∀n.
2. {bn} is a decreasing sequence
3. $\lim_{n \to ∞} b_n = 0$, then, the alternating series ∑an converges.

Proof (Credits: Alternating Series Test. Wikipedia).

Without loss of generality we can assume that the series starts at n = 1, and that we are given a series of the form $\sum_{n=1}^\infty (-1)^{n-1}a_n = a_1 -a_2 + a_3 -a_4 +···$ (the case $\sum_{n=1}^\infty (-1)^{n}a_n$ follows by taking the negative) where $\lim_{n \to ∞} a_n = 0, a_n ≥ 0, a_n ≥ a_{n+1}$

Claim: $S_{2m+1} = \sum_{n=1}^{2m+1} (-1)^{n-1}a_n→ L, S_{2m} = \sum_{n=1}^{2m} (-1)^{n-1}a_n → L$ ⇒ The usual partial sum, $S_k = \sum_{n=1}^{k} (-1)^{n-1}a_n → L$ and we are done.

S2(m+1)+1 = S2m+3 = S2(m+1) (odd) -a2m+2 (even) +a2m+3 (odd) ≤[Recall $a_n ≥ a_{n+1} ⇒ a_{2m+2} ≥ a_{2m+3} ⇒ a_{2m+3} - a_{2m+2} ≤ 0$] S2m+1 ⇒ the odd partial sums decrease monotonically.

S2(m+1) = S2m +a2m+1 (odd) -a2m+2 (even) ≥[Recall $a_n ≥ a_{n+1} ⇒ a_{2m+1} ≥ a_{2m+2} ⇒ a_{2m+1} - a_{2m+2} ≥ 0$] S2m ⇒ the even partial sums increase monotonically.

By assumption an≥ 0 ⇒ S2m+1 -S2m = a2m+1 ≥ 0.

Putting all this fact together, a1 -a2 = S2 ≤[The even partial sum increase monotonically] ≤ S4 ···for any arbitrary m ··· ≤ S2m ≤[S2m+1 -S2m = a2m+1 ≥ 0 ⇒ S2m+1 ≥ S2m] S2m+1 ≤[the odd partial sums decrease monotonically] ≤ S2(m-1)+1 ≤ ··· ≤ S1 = a1.

Therefore, a1-a2 is a lower bound of the monotonically decreasing sequence S2m+1 ⇒ [Monotone convergence theorem. Let {an} be a sequence of real numbers. If {an} is increasing and bounded above, then it is convergent. If {an} is decreasing and bounded below, then it is convergent."] {S2m+1} converges, say {S2m+1} → L1. Similarly, a1 is an upper bound of the monotonically increasing sequence S2m ⇒ [Monotone convergence theorem] {S2m} converges, say {S2m} → L2

$\lim_{m \to ∞} (S_{2m+1}-S_{2m}) = \lim_{m \to ∞} a_{2m+1}$ =[By assumption, $\lim_{n \to ∞} a_n = 0$] = 0.

$\lim_{m \to ∞} (S_{2m+1}-S_{2m}) = \lim_{m \to ∞} S_{2m+1} -\lim_{m \to ∞}-S_{2m} = L_1-L_2 = 0⇒ L_1 = L_2∎$

# Solved exercises

• Determine convergence or divergence of the series $\sum_{n=1}^\infty \frac{(-1)^n}{n}$

$\sum_{n=1}^\infty \frac{(-1)^n}{n} = -1 + \frac{1}{2} +\frac{-1}{3}+\frac{1}{4}+···$

To apply the Alternating Series Test, an = (-1)nbn, $b_n = \frac{1}{n}$

1. $\lim_{n \to ∞} b_n = \lim_{n \to ∞}\frac{1}{n} = 0$
2. {bn} is a decreasing sequence, $\frac{1}{n+1} ≤ \frac{1}{n}$ ⇒[By the Alternating Series Test] $\sum_{n=1}^\infty \frac{(-1)^n}{n}$ is convergent.
• Determine convergence or divergence of the series $\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}$

$\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}} = \frac{-1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{-1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+···$

To apply the Alternating Series Test, an = (-1)nbn, $b_n = \frac{1}{\sqrt{n}}$

1. $\lim_{n \to ∞} b_n = \lim_{n \to ∞}\frac{1}{\sqrt{n}} = 0$
2. {bn} is a decreasing sequence. $\frac{1}{\sqrt{n+1}} ≤ \frac{1}{\sqrt{n}}$ ⇒[By the Alternating Series Test] $\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}$ is convergent.

$\sum_{n=1}^\infty |a_n| = \sum_{n=1}^\infty |\frac{(-1)^n}{\sqrt{n}}| = \sum_{n=1}^\infty \frac{1}{\sqrt{n}}$, this is a p-series where p = 12 < 1 ⇒[Theorem (p-series)] $\sum_{n=1}^\infty |a_n|$ diverges ⇒ $\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}$ is conditionally convergent.

• Determine convergence or divergence of the series $\sum_{n=1}^\infty (-1)^{n+1}·\frac{5n+3}{2n-7}$

To apply the Alternating Series Test, an = (-1)n+1bn, $b_n =\frac{5n+3}{2n-7}$

1. $\lim_{n \to ∞} b_n = \lim_{n \to ∞}\frac{5n+3}{2n-7} =[L’Hôpital, ∞/∞] = \frac{5}{2} ≠ 0$ ⇒ Since one of the conditions of the Alternating Series Test is not satisfied, we cannot conclude whether the series $\sum_{n=1}^\infty (-1)^{n+1}·\frac{5n+3}{2n-7}$ converges or diverges.
2. {bn} is a decreasing sequence.
• Determine convergence or divergence of the series $\sum_{n=1}^\infty (-1)^{n}·\frac{3n-1}{2n+1}$

To apply the Alternating Series Test, an = (-1)nbn, $b_n =\frac{3n-1}{2n+1}$

1. $\lim_{n \to ∞} b_n = \lim_{n \to ∞}\frac{3n-1}{2n+1} =[L’Hôpital, ∞/∞] = \frac{3}{2} ≠ 0$ ⇒ Since one of the conditions of the Alternating Series Test is not satisfied, we cannot conclude whether the series $\sum_{n=1}^\infty (-1)^{n}·\frac{3n-1}{2n+1}$ converges or diverges.
2. {bn} is a decreasing sequence.
• Determine convergence or divergence of the series $\sum_{n=1}^\infty (-1)^n\frac{n}{(n^2+2)}$

To apply the Alternating Series Test, an = (-1)nbn, $b_n =\frac{n}{(n^2+2)}$

1. $\lim_{n \to ∞} b_n = \lim_{n \to ∞}\frac{n}{(n^2+2)} =[L’Hôpital, ∞/∞] = \lim_{n \to ∞}\frac{1}{2n} = 0$.
2. {bn} is a decreasing sequence (it is left as an easy exercise), hence the series $\sum_{n=1}^\infty (-1)^{n}·\frac{3n-1}{2n+1}$ is convergent.
• Determine convergence or divergence of the series $\sum_{n=1}^\infty \frac{(-1)^{n}}{3ln(n)}$

To apply the Alternating Series Test, an = (-1)nbn, $b_n =\frac{1}{3ln(n)}$

1. $\lim_{n \to ∞} b_n = \lim_{n \to ∞}\frac{1}{3ln(n)} = 0$.
2. {bn} is a decreasing sequence. $\frac{1}{3ln(n+1)}≤ \frac{1}{3ln(n)}$ (because ln(x) is an increasing function) ⇒ $\sum_{n=1}^\infty \frac{(-1)^{n}}{3ln(n)}$ is convergent by the Alternating Series Test.
• Determine convergence or divergence of the series $\sum_{n=1}^\infty (-1)^{n}·\frac{ln(n)}{n}$

To apply the Alternating Series Test, an = (-1)nbn, $b_n =\frac{ln(n)}{n}$

1. $\lim_{n \to ∞} b_n = \lim_{n \to ∞}\frac{ln(n)}{n} =[L’Hôpital, ∞/∞] = \lim_{n \to ∞} \frac{\frac{1}{n}}{1} = \lim_{n \to ∞} \frac{1}{n} = 0$.
2. {bn} is a decreasing sequence. This is not true, b1=0 , b2≈0.347, b3≈0.366, and b4≈0.347 for at least the first 3 terms. b’n = $\frac{1-ln(n)}{n^2} ≤ 0$ (∀n≥ 3, actually, for all n > e) ⇒ the series $\sum_{n=3}^\infty (-1)^{n}·\frac{ln(n)}{n}$ is convergent ⇒[Adding two finite terms, namely a1 and a2, do not change the convergence] $\sum_{n=1}^\infty (-1)^{n}·\frac{ln(n)}{n}$ is convergent

# Bibliography

1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Calculus and Calculus 3e (Apex). Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn, and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. YouTube’s Andrew Misseldine: Calculus. College Algebra and Abstract Algebra.
8. MIT OpenCourseWare 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007.
9. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.
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