Which Of The Following Series Is Divergent

Which of the Following Series is Divergent

Understanding series convergence and divergence is fundamental in mathematical analysis. A series is essentially the sum of the terms of a sequence, and determining whether a series converges (approaches a finite limit) or diverges (does not approach a finite limit) is crucial in many areas of mathematics and its applications. In this article, we'll explore how to identify divergent series and examine common examples that demonstrate divergence.

Introduction to Series and Convergence

A series is the sum of the terms of an infinite sequence. Mathematically, if we have a sequence {a₁, a₂, a₃, ..., aₙ, ...

∑aₙ = a₁ + a₂ + a₃ + ... + aₙ + ...

The question of whether this sum approaches a finite value (converges) or grows without bound (diverges) is central to the study of infinite series. Divergent series are those that do not converge to a finite limit. So in practice, as we add more terms, the partial sums either grow infinitely large, oscillate without settling to a particular value, or behave in some other manner that prevents convergence Worth knowing..

People argue about this. Here's where I land on it And that's really what it comes down to..

Basic Tests for Divergence

Several tests can help determine if a series is divergent. Here are the most fundamental ones:

The nth Term Test for Divergence

The simplest test for divergence is the nth term test. If the limit of the terms aₙ as n approaches infinity is not zero, then the series ∑aₙ must diverge.

Mathematically, if lim(n→∞) aₙ ≠ 0, then ∑aₙ diverges.

Important note: If lim(n→∞) aₙ = 0, the test is inconclusive. The series might converge or diverge.

Comparison Tests

Comparison tests compare a series to another series with known convergence properties:

1. Direct Comparison Test: If 0 ≤ aₙ ≤ bₙ for all n, and ∑bₙ diverges, then ∑aₙ also diverges Surprisingly effective..

2. Limit Comparison Test: If lim(n→∞) (aₙ/bₙ) = L where 0 < L < ∞, then both series ∑aₙ and ∑bₙ either both converge or both diverge Small thing, real impact..

Common Examples of Divergent Series

Let's examine several classic examples of divergent series and understand why they diverge.

The Harmonic Series

The harmonic series is one of the most famous examples of a divergent series:

∑(1/n) = 1 + 1/2 + 1/3 + 1/4 + 1/5 + ...

At first glance, since the terms 1/n approach 0 as n increases, one might suspect the series converges. On the flip side, the harmonic series actually diverges, meaning its sum grows infinitely large.

To prove this, we can group the terms:

1 + (1/2) + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + ...

Each grouped pair is greater than 1/2:

• 1/3 + 1/4 > 1/4 + 1/4 = 1/2
• 1/5 + 1/6 + 1/7 + 1/8 > 1/8 + 1/8 + 1/8 + 1/8 = 1/2

Since we can add infinitely many groups each greater than 1/2, the sum grows without bound And it works..

Geometric Series with |r| ≥ 1

A geometric series has the form:

∑arⁿ⁻¹ = a + ar + ar² + ar³ + .. Nothing fancy..

This series converges if |r| < 1 and diverges if |r| ≥ 1.

When |r| ≥ 1, the terms do not approach zero (except when r = 1 and a = 0, which is trivial), so by the nth term test, the series diverges That alone is useful..

For example:

• ∑2ⁿ = 2 + 4 + 8 + 16 + ... clearly diverges to infinity
• ∑(-1)ⁿ = 1 - 1 + 1 - 1 + ... oscillates and does not approach a finite limit

p-Series with p ≤ 1

A p-series has the form:

∑(1/nᵖ) = 1 + 1/2ᵖ + 1/3ᵖ + 1/4ᵖ + .. Not complicated — just consistent..

This series converges if p > 1 and diverges if p ≤ 1.

When p = 1, we get the harmonic series, which we've established diverges. For p < 1, the terms 1/nᵖ decrease more slowly than 1/n, so the series also diverges.

Series with Terms Not Approaching Zero

Any series where the terms do not approach zero as n approaches infinity must diverge by the nth term test.

For example:

• ∑(n/(n+1)) = 1/2 + 2/3 + 3/4 + 4/5 + ... The terms approach 1, not 0, so the series diverges.

And - ∑sin(n) = sin(1) + sin(2) + sin(3) + ... The terms oscillate between -1 and 1 without approaching 0, so the series diverges The details matter here..

Advanced Tests for Divergence

For more complex series, additional tests may be necessary:

Ratio Test

The ratio test examines the limit of the ratio of consecutive terms:

lim(n→∞) |aₙ₊₁/aₙ| = L

• If L > 1, the series diverges
• If L < 1, the series converges
• If L = 1, the test is inconclusive

Root Test

The root test examines the limit of the nth root of the absolute value of terms:

lim(n→∞) |aₙ|^(1/n) = L

• If L > 1, the series diverges
• If L < 1, the series converges
• If L = 1, the test is inconclusive

Integral Test

For series with positive, decreasing terms, we can compare the

Integral Test

When aseries has positive, decreasing terms (a_n) that can be represented by a function (f(x)) (i.e., (f(n)=a_n) for all integers (n\ge 1) and (f) is continuous, positive, and decreasing on ([1,\infty))), we may compare the series to an improper integral:

[ \sum_{n=1}^{\infty} a_n \quad\text{and}\quad \int_{1}^{\infty} f(x),dx . ]

The Integral Test states:

• If (\displaystyle\int_{1}^{\infty} f(x),dx) converges, then (\displaystyle\sum_{n=1}^{\infty} a_n) converges.
• If (\displaystyle\int_{1}^{\infty} f(x),dx) diverges, then (\displaystyle\sum_{n=1}^{\infty} a_n) diverges.

Example. Consider the p‑series (\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^{p}}) with (p>0).
Take (f(x)=\frac{1}{x^{p}}). Then

[ \int_{1}^{\infty}\frac{dx}{x^{p}}= \begin{cases} \displaystyle\frac{1}{p-1}, & p>1,\[6pt] \displaystyle\infty, & p\le 1. \end{cases} ]

Hence the p‑series converges exactly when (p>1), which matches the earlier result obtained by elementary comparison.

Direct Comparison Test

If we have two series with non‑negative terms, (\sum a_n) and (\sum b_n), and there exists a constant (C>0) such that (0\le a_n\le C,b_n) for all sufficiently large (n), then:

• If (\sum b_n) converges, so does (\sum a_n).
• If (\sum a_n) diverges, so does (\sum b_n).

This test is especially handy when we can bound a “complicated” series by a simpler one whose behavior we already know.

Example.
[\sum_{n=2}^{\infty}\frac{1}{n(\ln n)^{2}} ] converges because (\frac{1}{n(\ln n)^{2}}\le \frac{1}{n^{1.5}}) for all (n\ge 3), and (\sum 1/n^{1.5}) is a convergent p‑series Surprisingly effective..

Limit Comparison Test

When direct comparison is cumbersome, the limit comparison test offers a more flexible approach. Given two series (\sum a_n) and (\sum b_n) with (a_n,b_n>0), compute

[ L=\lim_{n\to\infty}\frac{a_n}{b_n}. ]

If (0<L<\infty) (i.Here's the thing — e. , the limit exists and is a positive finite number), then both series either converge or diverge together And that's really what it comes down to..

Example.
To study (\displaystyle\sum_{n=1}^{\infty}\frac{2n+3}{n^{2}+1}), compare it with (\displaystyle\sum_{n=1}^{\infty}\frac{1}{n}).
Compute

[ L=\lim_{n\to\infty}\frac{(2n+3)/(n^{2}+1)}{1/n}= \lim_{n\to\infty}\frac{2n+3}{n^{2}+1}\cdot n =\lim_{n\to\infty}\frac{2n^{2}+3n}{n^{2}+1}=2. ]

Since (0<L=2<\infty) and the harmonic series diverges, the given series also diverges Turns out it matters..

Alternating Series Test (Leibniz’s Test)

An alternating series has the form (\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1}b_n) with (b_n\ge0). The series converges if:

1. (b_{n+1}\le b_n) for all sufficiently large (n) (the terms are eventually monotone decreasing), and
2. (\displaystyle\lim_{n\to\infty} b_n = 0).

If these conditions are met, the series converges, though not necessarily absolutely Most people skip this — try not to..

Example.
[ \sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{\sqrt{n}} ] converges by the Alternating Series Test because (\frac{1}{\sqrt{n}}) decreases to 0, even though (\sum \frac{1}{\sqrt{n}}) diverges (a p‑series with (p=\tfrac12\le1)).

Absolute and Conditional Convergence

A series (\sum a_n) is said to converge absolutely if (\sum |a_n|) converges. If (\sum a_n) converges but (\sum |a_n|) diverges, the convergence is conditional Easy to understand, harder to ignore. Surprisingly effective..

Absolute convergence guarantees many useful properties (e.So g. , rearrangements do not affect the sum), whereas conditional convergence can be delicate.

Example.
The alternating harmonic series (\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{n}) converges conditionally (by the Alternating Series Test) but does not converge absolutely, since (\sum \frac{1}{n}) diverges.

About the Ra —tio Test is particularly useful for series with terms involving factorials, exponentials, or polynomials. }{(n+1)^{n+1}} \cdot \frac{n^n}{n!} = \frac{n^n}{(n+1)^n} \cdot \frac{1}{n+1} ). In real terms, here, ( \frac{a_{n+1}}{a_n} = \frac{(n+1)! If ( L < 1 ), the series converges absolutely; if ( L > 1 ), it diverges; and if ( L = 1 ), the test is inconclusive. Here's one way to look at it: consider ( \sum_{n=1}^{\infty} \frac{n!}{n^n} ). It involves computing the limit ( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ). As ( n \to \infty ), this simplifies to ( \lim_{n \to \infty} \frac{1}{e(n+1)} = 0 ), confirming absolute convergence That's the part that actually makes a difference..

The Root Test, which evaluates ( L = \lim_{n \to \infty} \sqrt[n]{|a_n|} ), is effective for series with terms raised to the ( n )-th power. If ( L < 1 ), the series converges absolutely; if ( L > 1 ), it diverges; and if ( L = 1 ), the test is inconclusive. To give you an idea, ( \sum_{n=1}^{\infty} \left( \frac{3n^2 + 2}{5n^2 + 7} \right)^n ) yields ( L = \lim_{n \to \infty} \frac{3n^2 + 2}{5n^2 + 7} = \frac{3}{5} < 1 ), so the series converges absolutely.

The Integral Test connects series convergence to improper integrals. For a continuous, positive, decreasing function ( f(n) = a_n ), the series ( \sum_{n=1}^{\infty} a_n ) converges if and only if ( \int_{1}^{\infty} f(x) , dx ) converges. Here's one way to look at it: ( \sum_{n=1}^{\infty} \frac{1}{n (\ln n)^2} ) can be analyzed via the integral ( \int_{2}^{\infty} \frac{1}{x (\ln x)^2} , dx ). Substituting ( u = \ln x ), the integral becomes ( \int_{\ln 2}^{\infty} \frac{1}{u^2} , du ), which converges. Thus, the series converges.

The Comparison Test and Limit Comparison Test are foundational for analyzing series with positive terms. In real terms, the Comparison Test requires bounding ( a_n ) by a known convergent or divergent series, while the Limit Comparison Test uses the ratio ( \lim_{n \to \infty} \frac{a_n}{b_n} ). Take this: ( \sum_{n=1}^{\infty} \frac{2n + 3}{n^2 + 1} ) is compared to ( \sum \frac{1}{n} ), yielding ( L = 2 ), so both series diverge.

The Alternating Series Test applies to series of the form ( \sum (-1)^{n-1} b_n ) with ( b_n \geq 0 ). So convergence requires ( b_n ) to be eventually decreasing and approach zero. Here's one way to look at it: ( \sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{\sqrt{n}} ) converges because ( \frac{1}{\sqrt{n}} ) decreases to zero, even though ( \sum \frac{1}{\sqrt{n}} ) diverges That's the whole idea..

Absolute and conditional convergence distinguish between series that converge on their own terms (( \sum |a_n| ) converges) and those that only converge conditionally (( \sum a_n ) converges but ( \sum |a_n| ) diverges). The alternating harmonic series ( \sum_{n=1}^{\infty} (-1)^{n-1} \frac{1}{n} ) converges conditionally, as its absolute series ( \sum \frac{1}{n} ) diverges.

This changes depending on context. Keep that in mind.

For series with mixed positive and negative terms, the Comparison Test is not directly applicable, but absolute convergence can be tested via ( \sum |a_n| ). Here's one way to look at it: ( \sum_{n=1}^{\infty} (-1)^n \frac{n}{n^2 + 1} ) converges absolutely because ( \sum \frac{n}{n^2 + 1} ) is comparable to ( \sum \frac{1}{n} ), which diverges, but ( \sum \left| (-1)^n \frac{n}{n^2 + 1} \right| = \sum \frac{n}{n^2 + 1} ) diverges, so the original series converges conditionally Simple as that..

Pulling it all together, these tests provide a solid framework for analyzing series convergence. In practice, absolute convergence ensures stability under rearrangement, while conditional convergence requires careful handling. The Comparison and Limit Comparison Tests are ideal for positive-term series, the Ratio and Root Tests handle factorial or exponential terms, the Integral Test links series to integrals, and the Alternating Series Test addresses conditional convergence. Mastery of these tools is essential for solving complex series problems in analysis and applied mathematics Easy to understand, harder to ignore..

\boxed{\text{These tests collectively enable the determination of convergence for a wide range of series, emphasizing the importance of choosing

[ \boxed{\text{These tests collectively enable the determination of convergence for a wide range of series, emphasizing the importance of choosing the appropriate tool for the structure of the terms.}} ]

5. The Ratio and Root Tests in Practice

While the Ratio and Root Tests were mentioned briefly earlier, their utility shines when dealing with factorials, powers, or combinations thereof Small thing, real impact..

5.1 Ratio Test

Given a series (\sum a_n) with (a_n \neq 0) for large (n), define

[ L=\lim_{n\to\infty}\Bigl|\frac{a_{n+1}}{a_n}\Bigr|. ]

• If (L<1), the series converges absolutely.
• If (L>1) (including (L=\infty)), the series diverges.
• If (L=1), the test is inconclusive.

Example:
[ \sum_{n=1}^{\infty}\frac{n!}{3^n} ] Here

[ \Bigl|\frac{a_{n+1}}{a_n}\Bigr|=\frac{(n+1)!}{3^{,n+1}}\cdot\frac{3^{,n}}{n!}= \frac{n+1}{3}\xrightarrow[n\to\infty]{}\infty, ]

so (L>1) and the series diverges.

5.2 Root Test

For (\sum a_n) define

[ L=\limsup_{n\to\infty}\sqrt[n]{|a_n|}. ]

The conclusions are identical to the Ratio Test. The Root Test is particularly convenient when each term is raised to the (n)‑th power.

Example:
[ \sum_{n=1}^{\infty}\Bigl(\frac{2n+1}{3n-2}\Bigr)^{!n}. ] We compute

[ \sqrt[n]{|a_n|}= \frac{2n+1}{3n-2}\xrightarrow[n\to\infty]{}\frac{2}{3}<1, ]

hence the series converges absolutely Small thing, real impact..

6. The Integral Test Revisited

The Integral Test not only provides a convergence criterion but also yields useful error estimates for series of positive, decreasing terms. If

[ f(x)=a_x,\qquad f\text{ decreasing on }[N,\infty), ]

then

[ \int_{N+1}^{\infty} f(x),dx \le \sum_{n=N+1}^{\infty} a_n \le \int_{N}^{\infty} f(x),dx . ]

This means the remainder after (N) terms satisfies

[ R_N = \sum_{n=N+1}^{\infty} a_n \le \int_{N}^{\infty} f(x),dx . ]

Example: Approximate (\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}) by its first three terms.
Take (f(x)=1/x^2). Then

[ \int_{3}^{\infty}\frac{dx}{x^{2}} = \frac{1}{3}, ]

so the truncation error is less than (1/3). In fact, the exact remainder is (\pi^2/6 - (1+1/4+1/9) \approx 0.105), confirming the bound Worth keeping that in mind..

7. Alternating Series Error Estimate

When an alternating series satisfies the hypotheses of the Alternating Series Test, the absolute error after truncating after (N) terms is bounded by the first omitted term:

[ \bigl|S - S_N \bigr| \le b_{N+1}. ]

This simple estimate is particularly useful in numerical work Easy to understand, harder to ignore..

Example: Approximate (\displaystyle\ln(2)=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{n}) using the first five terms.

[ S_5 = 1-\frac12+\frac13-\frac14+\frac15 = 0.78333\ldots ]

The next term in absolute value is (b_6 = 1/6 \approx 0.That's why 78333 \pm 0. 1667); thus the true value lies within (0.Indeed, (\ln 2 \approx 0.So 1667). 6931), well inside the interval Simple, but easy to overlook. That alone is useful..

8. Conditional Convergence and Rearrangement

A striking consequence of conditional convergence is the Riemann rearrangement theorem: if a series converges conditionally, its terms can be rearranged to converge to any prescribed real number, or even to diverge. This phenomenon underscores why absolute convergence is a stronger, more “stable” property.

Illustration:

Consider the alternating harmonic series

[ S = \sum_{n=1}^{\infty}(-1)^{n-1}\frac{1}{n}= \ln 2. ]

By grouping two positive terms followed by one negative term, we obtain a new series that converges to a larger limit (approximately (1.1)). Even so, by suitably biasing the selection of positive versus negative terms, the partial sums can be forced toward any target value. This flexibility disappears once the series is absolutely convergent, because absolute convergence guarantees that any rearrangement yields the same sum.

9. Power Series and Radius of Convergence

Power series (\displaystyle\sum_{n=0}^{\infty}c_n (x-a)^n) are a natural arena where the Ratio and Root Tests combine to give the radius of convergence (R) It's one of those things that adds up..

• Using the Ratio Test:

[ R = \lim_{n\to\infty}\Bigl|\frac{c_n}{c_{n+1}}\Bigr|. ]

• Using the Root Test:

[ R = \frac{1}{\displaystyle\limsup_{n\to\infty}\sqrt[n]{|c_n|}}. ]

Inside the interval ((a-R, a+R)) the series converges absolutely; outside it diverges. At the endpoints (x=a\pm R) one must test convergence separately, often employing the tests discussed earlier Simple as that..

Example:

[ \sum_{n=0}^{\infty}\frac{(x-2)^n}{n!} ]

Applying the Ratio Test:

[ \Bigl|\frac{a_{n+1}}{a_n}\Bigr| = \frac{|x-2|^{n+1}}{(n+1)!}\cdot\frac{n!}{|x-2|^{n}} = \frac{|x-2|}{n+1}\xrightarrow[n\to\infty]{}0. ]

Since the limit is (0<1) for every finite (x), the radius of convergence is (R=\infty); the series defines the exponential function (e^{x-2}).

10. A Unified Decision Tree

When faced with a new series (\sum a_n), a systematic approach can save time:

1. Check sign – if all (a_n\ge0), start with Comparison, Limit Comparison, or Integral Test.
2. Identify factorials, exponentials, or powers of (n) – apply Ratio or Root Test.
3. Detect alternating signs – verify monotonic decrease of (|a_n|) and limit zero; use Alternating Series Test.
4. If convergence is established, test absolute convergence by examining (\sum|a_n|).
5. For power series, compute the radius of convergence via Ratio/Root Test, then treat endpoints individually.

Following this decision tree ensures that the most efficient test is applied first, reducing redundant calculations.

Conclusion

The landscape of infinite series is rich, but the array of convergence tests—Comparison, Limit Comparison, Integral, Ratio, Root, Alternating Series, and the concepts of absolute versus conditional convergence—forms a cohesive toolkit. Day to day, each test exploits a different structural feature of the terms: size comparison, growth rate, sign pattern, or relationship to an integral. Mastery of these methods not only enables the rigorous determination of whether a series converges, but also provides quantitative error bounds, insight into rearrangement phenomena, and the foundation for more advanced topics such as power series and Fourier analysis.

By internalizing the criteria and practicing the decision‑making process outlined above, students and practitioners can approach any series with confidence, selecting the optimal test and interpreting the result within the broader context of mathematical analysis The details matter here..