Alternating Series

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Mathematical Series Review

A finite series is given by all the terms of a finite sequence added together. For example, the sum of the sequence $\{3, 5, 7, \dots, 21\}$ is written as: $\sum_{k=1}^{10} (2k+1) = 120$.

An infinite series is the sum of an infinite sequence of numbers. It is represented in the form $\sum_{n=1}^\infty a_n = a_1 + a_2 + a_3 + \dots$ where $a_n$ represents the terms of the sequence, and $n$ is the index that ranges from $1$ to infinity.

Convergence and Divergence

Definition. A series is convergent (or converges) if the sequence of its partial sums tends to a limit. More precisely, if we define the $n$-th partial sum as $S_n = \sum_{k=1}^n a_k = a_1 + a_2 + \dots + a_n$, the series converges if: $L = \lim_{n \to \infty} S_n$ exists and is a finite number.

Formally, this means there exists a number $L$ such that for every arbitrarily small positive number $\varepsilon$, there is a sufficiently large integer $N$, such that for all $n \geq N$, $\vert{}S_n - L\vert{} < \varepsilon$. If the series is convergent, the number $L$ is called the sum of the series.

On the contrary, any series that is not convergent (where the limit of $S_n$ does not exist or approaches infinity) is said to be divergent (or to diverge).

Convergence Tests

Divergence Test ($n$-th Term Test). If $\lim_{n \to \infty} a_n \neq 0$, then $\sum_{n=1}^\infty a_n$ diverges.
If the limit is 0, this test is inconclusive; the series might converge or diverge.

Integral Test. If $f(x)$ is a positive, continuous, and decreasing function on the interval $[1, \infty)$, then the series $\sum_{n=1}^\infty f(n)$ converges if and only if the improper integral $\int_{1}^{\infty} f(x) dx$ converges.

Direct Comparison Test. Let $\{a_n\}$ and $\{b_n\}$ be positive sequences where $a_n \leq b_n$ for all $n \geq N$, for some integer $N$.

1. If the “larger” series $\sum_{n=1}^\infty b_n$ converges, then the “smaller” series $\sum_{n=1}^\infty a_n$ converges.
2. If the “smaller” series $\sum_{n=1}^\infty a_n$ diverges, then the “larger” series $\sum_{n=1}^\infty b_n$ diverges.

Absolute vs. Conditional Convergence

Absolute Convergence: An infinite series is said to converge absolutely if the absolute values of its terms form a convergent series (i.e., $\sum_{n=1}^\infty \vert{}a_n\vert{} = L$ for some real number $L$).
Theorem: If a series converges absolutely, it also converges in the normal sense.

Conditional Convergence: A series is conditionally convergent if it converges normally, 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 $a_n$ are positive or all $a_n$ are negative.

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

1. $|a_n|$ decreases monotonically, i.e., $|a_{n+1}| \leq |a_n|$.
2. $\lim_{n \to \infty} a_n = 0$.

Other authors express it as follows. Suppose that we have a series $\sum_{n=0}^\infty a_n$ and either $a_n = (-1)^nb_n$ or $a_n = (-1)^{n+1}b_n$, let’s also suppose that

1. $b_n$ positive, $b_n \gt 0, \forall n$
2. $\{ b_n \}$ is a decreasing sequence
3. $\lim_{n \to \infty} b_n = 0$, then, the alternating series $\sum_{n=0}^\infty a_n$ converges.

Proof of the Alternating Series Test

1. Without loss of generality, assume the series is $\sum_{n=1}^\infty (-1)^{n-1} b_n = b_1 - b_2 + b_3 - b_4 + \dots$ where $b_n \geq 0$, $b_n \geq b_{n+1}$, and $\lim_{n \to \infty} b_n = 0$.
2. We can examine the partial sums by splitting them into odd and even terms.
3. Odd partial sums decrease monotonically.
   $S_{2m+3} = S_{2m+1} - b_{2m+2} + b_{2m+3}$. Because the terms are decreasing ($b_{2m+2} \geq b_{2m+3}$), the quantity $(-b_{2m+2} + b_{2m+3}) \leq 0$. Therefore, $S_{2m+3} \leq S_{2m+1}$.
4. Even partial sums increase monotonically.
   $S_{2m+2} = S_{2m} + b_{2m+1} - b_{2m+2}$. Because $b_{2m+1} \geq b_{2m+2}$, the quantity $(b_{2m+1} - b_{2m+2}) \geq 0$. Therefore, $S_{2m+2} \geq S_{2m}$.
5. Since $b_n \geq 0$, we know $S_{2m+1} - S_{2m} = b_{2m+1} \geq 0$, meaning every odd sum is greater than the even sum right before it.
6. Putting it all together: $S_2 \leq S_4 \leq \dots \leq S_{2m} \leq S_{2m+1} \leq \dots \leq S_3 \leq S_1$.
7. By the Monotone Convergence Theorem, since the odd sums decrease and are bounded below (by $S_2$), they converge to a limit $L_1$. Similarly, the even sums increase and are bounded above (by $S_1$), converging to a limit $L_2$.
8. Finally, let’s find the difference between these limits: $\lim_{m \to \infty} (S_{2m+1} - S_{2m}) = \lim_{m \to \infty} b_{2m+1} =[\text{By assumption }, \lim_{n \to \infty} b_n = 0] 0$
   $\lim_{m \to \infty} (S_{2m+1} - S_{2m}) = \lim_{m \to \infty} S_{2m+1} -\lim_{m \to \infty}-S_{2m} = L_1-L_2 = 0 \implies L_1 = L_2 \blacksquare$

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, $a_n = (-1)^n b_n, b_n = \frac{1}{n}$

1. Limit condition: $\lim_{n \to \infty} b_n = \lim_{n \to \infty}\frac{1}{n} = 0$
2. $\{ b_n \}$ is a decreasing sequence, $\frac{1}{n+1} \leq \frac{1}{n}$.
3. Conclusion: 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, $a_n = (-1)^n b_n, b_n = = \frac{1}{\sqrt{n}}$

1. Limit condition: $\lim_{n \to \infty} b_n = \lim_{n \to \infty}\frac{1}{\sqrt{n}} = 0$
2. $\{ b_n \}$ is a decreasing sequence. $\frac{1}{\sqrt{n+1}} \leq \frac{1}{\sqrt{n}}$.
3. By the Alternating Series Test, $\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}$ is convergent.
4. However, if we take the absolute value, $\sum_{n=1}^\infty |a_n| = \sum_{n=1}^\infty |\frac{(-1)^n}{\sqrt{n}}| = \sum_{n=1}^\infty \frac{1}{\sqrt{n}}$, this is a p-series where $p = \frac12 \lt 1 \implies$ [Theorem (p-series)] $\sum_{n=1}^\infty |a_n|$ diverges.
5. Conclusion. $\sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n}}$ 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, $a_n = (-1)^{n+1} b_n, b_n = \frac{5n+3}{2n-7}$

1. Limit condition: $\lim_{n \to \infty} b_n = \lim_{n \to \infty}\frac{5n+3}{2n-7} =[L'Hôpital, \infty/ \infty] = \frac{5}{2} \neq 0$ ⇒ 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. Because the terms $b_n$ approach $\frac{5}{2}$, the alternating sequence $a_n$ approaches both $\frac{5}{2}$ and $-\frac{5}{2}$ as $n$ goes to infinity.
3. Since $\lim_{n \to \infty} a_n \neq 0$, the series diverges by the Divergence Test. The AST is inconclusive, but the Divergence Test gives us a definitive answer.

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

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

1. Limit condition: $\lim_{n \to \infty} b_n = \lim_{n \to \infty}\frac{3n-1}{2n+1} =[L'Hôpital, \infty/\infty] = \frac{3}{2} \neq 0$ ⇒ 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. Just like the previous example, the limit of the general term does not equal zero. The series diverges by the Divergence Test.

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

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

1. Limit condition: $\lim_{n \to \infty} b_n = \lim_{n \to \infty}\frac{n}{(n^2+2)} =[L'Hôpital, \infty/\infty] = \lim_{n \to \infty}\frac{1}{2n} = 0$.
2. $\{ b_n \}$ is a decreasing sequence. Using the quotient rule:
   $f'(x) = \frac{1(x^2+2) - x(2x)}{(x^2+2)^2} = \frac{2 - x^2}{(x^2+2)^2}$
   For $x \geq \sqrt{2}$, $f'(x) < 0$, which means the sequence $b_n$ is strictly decreasing for $n \geq 2$.
3. Conclusion: By the Alternating Series Test, $\sum_{n=1}^\infty (-1)^{n}·\frac{3n-1}{2n+1}$ is convergent.

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

1. Important Domain Note: At $n = 1$, $\ln(1) = 0$, which makes the first term undefined ($\frac{1}{0}$). Therefore, this series must start at $n = 2$.
   To apply the Alternating Series Test (AST), write $a_n = (-1)^n b_n$ where $b_n = \frac{1}{3\ln(n)}$ for $n \ge 2$.
2. Check the Limit. $\lim_{n \to \infty} b_n = \lim_{n \to \infty} \frac{1}{3\ln(n)} = 0$
   Because $\ln(n) \to \infty$ as $n \to \infty$, the denominator grows without bound, making the limit $0$.
3. Check Monotonicity (Decreasing Terms). We check if $b_{n+1} \le b_n$ for all $n \ge 2$:
   Since $\ln(x)$ is a strictly increasing function for $x > 0$: $n < n+1 \implies \ln(n) < \ln(n+1) \implies 3\ln(n) < 3\ln(n+1)$
   Taking the reciprocal flips the inequality: $\frac{1}{3\ln(n+1)} < \frac{1}{3\ln(n)} \implies b_{n+1} < b_n$
   Thus, $\{b_n\}$ is strictly decreasing for all $n \ge 2$.
4. Conclusion. By the Alternating Series Test, the series $\sum_{n=2}^\infty \frac{(-1)^{n}}{3\ln(n)}$ converges.
5. Is it Absolutely or Conditionally Convergent?
   Let’s test absolute convergence: $\sum_{n=2}^\infty \left\vert{} \frac{(-1)^n}{3\ln(n)} \right\vert{} = \sum_{n=2}^\infty \frac{1}{3\ln(n)}$.
   For $n \ge 2$, we know $n > \ln(n)$, which implies $\frac{1}{\ln(n)} > \frac{1}{n}$.
   Since the harmonic series $\sum_{n=2}^\infty \frac{1}{3n}$ diverges ($p$-series with $p=1$), by the Direct Comparison Test, $\sum_{n=2}^\infty \frac{1}{3\ln(n)}$ also diverges.
   $\therefore$ The series is conditionally convergent.

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

1. To apply the Alternating Series Test, write $a_n = (-1)^n b_n$ where $b_n = \frac{\ln(n)}{n}$ for $n \ge 1$.
2. Check the Limit.
   Since $\lim_{n \to \infty} \frac{\ln(n)}{n}$ is an indeterminate form of type $\left[\frac{\infty}{\infty}\right]$, apply L’Hôpital’s Rule: $\lim_{n \to \infty} b_n = \lim_{x \to \infty} \frac{\ln(x)}{x} \overset{\text{L'H}}{=} \lim_{x \to \infty} \frac{\frac{1}{x}}{1} = \lim_{x \to \infty} \frac{1}{x} = 0$
3. Check Monotonicity (Eventual Monotonicity). Evaluating the first few terms of $b_n$: $b_1 = \frac{\ln(1)}{1} = 0$, $b_2 = \frac{\ln(2)}{2} \approx 0.347$, $b_3 = \frac{\ln(3)}{3} \approx 0.366$, $b_4 = \frac{\ln(4)}{4} \approx 0.347$, etc.
   Notice that $b_n$ increases from $n = 1$ to $n = 3$, but then begins to decrease!
   To prove where it strictly decreases, differentiate $f(x) = \frac{\ln(x)}{x}$ using the Quotient Rule:
   $f'(x) = \frac{\frac{1}{x} \cdot x - \ln(x) \cdot 1}{x^2} = \frac{1 - \ln(x)}{x^2}$
   For $f'(x) < 0$, we need the numerator to be negative: $1 - \ln(x) < 0 \implies \ln(x) > 1 \implies x > e \approx 2.718$
   Since $e \approx 2.718$, $f'(x) < 0$ for all $x > e$. For integers, this means $b_n$ is strictly decreasing for all $n \ge 3$.
   The Eventual Monotonicity Rule: The Alternating Series Test does not require $b_n$ to decrease from the very first term ($n=1$). It only requires $b_n$ to be monotonically decreasing eventually (i.e., for $n \ge N$).
   Because adding or subtracting a finite number of terms (here, $a_1$ and $a_2$) has no effect on infinite convergence: $\sum_{n=1}^\infty (-1)^n \frac{\ln(n)}{n} = a_1 + a_2 + \sum_{n=3}^\infty (-1)^n \frac{\ln(n)}{n}$
4. Conclusion. Since the tail $\sum_{n=3}^\infty (-1)^n \frac{\ln(n)}{n}$ satisfies both conditions of the AST, it converges. Adding the finite terms $a_1$ and $a_2$ yields a finite sum. Therefore, the full series converges.
5. Is it Absolutely or Conditionally Convergent?
   Test absolute convergence: $\sum_{n=1}^\infty \left\vert{} (-1)^n \frac{\ln(n)}{n} \right\vert{} = \sum_{n=1}^\infty \frac{\ln(n)}{n}$.
   For $n \ge 3$, $\ln(n) > 1$, which gives: $\frac{\ln(n)}{n} > \frac{1}{n}$
   Since $\sum_{n=3}^\infty \frac{1}{n}$ diverges (Harmonic Series), by the Direct Comparison Test, $\sum_{n=1}^\infty \frac{\ln(n)}{n}$ diverges.
   $\therefore$ The series is conditionally convergent.

Bibliography

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