Testing for Convergence: The Divergence and Integral Tests

There seemed to be some correlation between devotion to God and a misguided zeal for marshmallow, David Sedaris.

Infinite Series. Convergence and Divergence.

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

A series is convergent (or converges) if the sequence of its partial sums tends to a finite limit, that is, $l = \lim_{n \to \infty} S_n = \lim_{n \to \infty} \sum_{k=1}^n a_k$ exists and is a finite number.

The $n$-th partial sum $S_n$ is the sum of the first $n$ terms: $S_n = \sum_{k=1}^n a_k = a_1 + a_2 + \dots + a_n$. If the limit $L = \lim_{n \to \infty} S_n$ exists and is a finite number, the series converges, and $L$ is called the sum of the series. Formally, this means for any arbitrarily small positive number $\epsilon$, there is a sufficiently large $N$ such that for all $n \geq N$, $\vert{}S_n - L\vert{} < \epsilon$.

If this limit does not exist (e.g., it goes to infinity or oscillates forever), the series is said to be divergent.

Testing for Convergence: The Divergence and Integral Tests

The Necessary Condition for Convergence (The Divergence Test)

Consider the harmonic series $\sum_{n=1}^\infty \frac{1}{n}$. Let’s compare it to the integral of the related function $f(x) = \frac{1}{x}$.

If we use upper Riemann sums with a width of $\Delta x = 1$ starting at $x=1$, the area of the rectangles is: $1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{N-1}$

This is exactly the $N$-th partial sum of our series, minus the last term. We can see geometrically that the curve $y = \frac{1}{x}$ sits underneath these rectangles. Therefore, the area under the curve is strictly less than the area of the rectangles: $\int_{1}^{N} \frac{1}{x} dx < 1 + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{N-1} < S_N$

We can easily evaluate the integral: $\int_{1}^{N} \frac{1}{x} dx = \ln(x) \bigg\vert{}_{1}^{N} = \ln(N) - \ln(1) = \ln(N)$. So, we know that $\ln(N) < S_N$ (Figure 2.1).

Because $\lim_{N \to \infty} \ln(N) = \infty$, it forces the partial sums $S_N$ to also go to infinity. This proves the harmonic series $\sum_{n=1}^\infty \frac{1}{n}$ diverges.

Taking the lower Riemann Sum with a width of $\Delta x = 1$ starting at $x=1$, the area of the rectangles is: $\int_{1}^{n} \frac{dx}{x} > \frac{1}{2} + \frac{1}{3} + ··· + \frac{1}{N} = S_N -1 \implies \ln(N) \gt \int_{1}^{n} \frac{dx}{x} > S_N -1 \implies \ln(N) < S_N < \ln(N) + 1$

This leads us to a fundamental theorem.

The Necessary Condition for Convergence

If a series $\sum_{n=1}^\infty a_n$ converges, then the limit of its terms must be zero: $\lim_{n \to \infty} a_n = 0$.

Proof:

Let $S_n$ be the $n$-th partial sum. We know that the $n$-th term is the difference between the last two partial sums: $a_n = S_n - S_{n-1}$.

If the series converges, its sequence of partial sums converges to some limit $L$. Thus, $\lim_{n \to \infty} S_n = L$, and consequently, $\lim_{n \to \infty} S_{n-1}$ is also $L$.

Taking the limit of the sequence terms: $\lim_{n \to \infty} a_n = \lim_{n \to \infty} (S_n - S_{n-1}) = \lim_{n \to \infty} S_n - \lim_{n \to \infty} S_{n-1} = L - L = 0$

The contrapositive of this theorem is one of the most useful tools in calculus.

The Divergence Test (n-th Term Test). If $\lim_{n \to \infty} a_n \neq 0$, then the series $\sum_{n=1}^\infty a_n$ diverges.

💀 A Crucial Warning!

The Divergence Test can only tell you if a series diverges. It cannot prove convergence. If the limit of the terms goes to zero, the test is inconclusive.

For example, $\lim_{n \to \infty} \frac{1}{n} = 0$ and $\lim_{n \to \infty} \frac{1}{n^2} = 0$. Both pass the necessary condition, yet we just proved the first series diverges, while the second series converges. Zero limit is necessary, but not sufficient.

Solved Examples: The Divergence Test

• $\sum_{n=1}^\infty \frac{4n+1}{5n-7}$
  Take the limit of the terms: $\lim_{n \to \infty} \frac{4n+1}{5n-7} =[\text{Use L'Hôpital's rule or simply divide top and bottom by } n] \frac{4}{5}$ Since $\frac{4}{5} \neq 0$, the series diverges by the Divergence Test.

• $\sum_{n=1}^\infty 3^n$
  Take the limit of the terms: $\lim_{n \to \infty} 3^n = \infty$
  Since $\infty \neq 0$, the series diverges by the Divergence Test.

• $\sum_{n=1}^\infty \cos\left(\frac{1}{n^2}\right)$
  As $n \to \infty$, the inside fraction $\frac{1}{n^2} \to 0$.
  Therefore: $\lim_{n \to \infty} \cos\left(\frac{1}{n^2}\right) = \cos(0) = 1 \ne 0$
  Since $1 \neq 0$, the series diverges by the Divergence Test.

The Integral Comparison Test

Sometimes, a series behaves exactly like its continuous counterpart.

The Integral Test. If $f(x)$ is a positive, continuous, and decreasing function on the interval $[k, \infty)$, and $a_n = f(n)$, then the series $\sum_{n=k}^\infty a_n$ and the improper integral $\int_{k}^{\infty} f(x) dx$ share the same fate:

• If the integral converges, the series converges.
• If the integral diverges, the series diverges.

Furthermore, we can bound the error between the series and the integral: $\left\vert{} \sum_{n=1}^\infty f(n) - \int_{1}^{\infty} f(x) dx \right\vert{} < f(1)$

Proof Intuition:

Using Riemann sums, we can “sandwich” the integral between two variations of the series sum.

Upper rectangles: The sum starting at $n=1$ creates rectangles that overshoot the curve, meaning $\int_{1}^{\infty} f(x) dx < \sum_{n=1}^\infty f(n)$ (Figure 2.a). If the integral diverges to infinity, the larger series must also diverge.

Lower rectangles: By shifting the rectangles rightward (starting the sum at $n=2$), they fit entirely under the curve, meaning $\sum_{n=2}^\infty f(n) < \int_{1}^{\infty} f(x) dx$. If the integral converges to a finite number, the smaller series is bounded and must also converge (Figure 2.b., the areas of the rectangles is less than the area under f(x), but this summation starts with n = 2).

The error bound

1. Lower rectangles $\sum_{n=2}^\infty f(n) < \int_{1}^{\infty} f(x) dx \implies[\text{add f(1) to both sides}] \sum_{n=1}^\infty f(n) < f(1) + \int_1^\infty f(x)dx$. Rearranging: $\sum_{n=1}^\infty f(n) - \int_1^\infty f(x)dx < f(1).$
2. Upper rectangles. From $\int_{1}^{\infty} f(x) dx < \sum_{n=1}^\infty f(n)$. Or equivalently: $\sum_{n=1}^\infty f(n) - \int_1^\infty f(x)dx > 0.$
3. Combine: $0 \lt f(n) - \int_1^\infty f(x)dx \lt f(1)$, and since the difference is positive, $\boxed{\left\vert{} \sum_{n=1}^\infty f(n) - \int_{1}^{\infty} f(x) dx \right\vert{} < f(1)}$

• If the integral converges, then the series converges (by the inequalities and monotonicity).
• If the integral diverges, then the series diverges.
• Equivalently: they share the same fate.

If $\sum_{n=1}^\infty f(n)$ diverges, so does $\int_{1}^{∞} f(x)dx$ because $\sum_{n=1}^\infty f(n) < a_1 + \int_{1}^{∞} f(x)dx$ (i). If $\int_{1}^{∞} f(x)dx$ diverges, so does $\sum_{n=1}^\infty f(n)$ because $\int_{1}^{∞} f(x)dx < \sum_{n=1}^\infty f(n)$

If $\sum_{n=1}^\infty f(n)$ converges, so does $\int_{1}^{∞} f(x)dx$ because $\int_{1}^{∞} f(x)dx < \sum_{n=1}^\infty f(n)$. If $\int_{1}^{∞} f(x)dx$ converges, so does $\sum_{n=1}^\infty f(n)$ because $\sum_{n=1}^\infty f(n) < a_1 + \int_{1}^{∞} f(x)dx$

Solved Examples: The Integral Test

• $\sum_{n=1}^\infty \frac{2}{3n+5}$. Let $f(x) = \frac{2}{3x+5}$ on $[1, \infty)$.

1. First, we must check the conditions before applying the integral test.
   Positive. For all $x \geq 1$, the denominator is positive, so $f(x) > 0$.
2. Continuous. It is a rational function. A rational function is continuous for all x except those values that make a denominator 0 ↭ 3x + 5 = 0. The only discontinuity is at $x = -5/3$, which is outside our interval.
3. Decreasing. f’(x) = $(-2)(3x+1)^{-2}·3 = \frac{-6}{(3x+1)^2}< 0$. Since $f'(x) < 0$ for all $x$, the function is strictly decreasing.
4. Evaluate the Improper Integral $\int_{1}^{\infty} \frac{2}{3x+5}dx = $[Use U-substitution, 3x + 5 = u ⇒ 3dx = du] $\frac{2}{3}\int_{1}^{∞} \frac{du}{u} = \lim_{b \to \infty} \frac{2}{3}\int_{1}^{b} \frac{du}{u} = \lim_{b \to ∞} \frac{2}{3}·ln|u| = \lim_{b \to ∞} \frac{2}{3}·ln|3x+5|\bigg|_{1}^{b}=$

$ \lim_{b \to \infty} \frac{2}{3} \left( \ln(3b+5) - \ln(8) \right) = \infty$. Since the integral evaluates to infinity (diverges), the series diverges by the Integral Test.

• $\sum_{n=1}^\infty \frac{1}{2^n}$

Step 1: Verify the 3 Conditions Let $f(x) = \frac{1}{2^x} = 2^{-x}$ on $[1, \infty)$.

1. Positive. Exponential functions are always positive.
2. Continuous. It is a rational function. A rational function is continuous for all x except those values that make a denominator 0, and $2^x \ne 0, \forall x \ge 1$.
3. Decreasing. $f'(x) = -2^{-x}\ln(2) = -\frac{\ln(2)}{2^x}$. Since $\ln(2)$ and $2^x$ are positive, the derivative is negative, meaning the function is decreasing.

Step 2: Evaluate the Improper Integral

Recall the integration rule: $\int a^x dx = \frac{a^x}{\ln(a)}$.

$\int_{1}^{\infty} 2^{-x} dx = \lim_{b \to \infty} \left[ \frac{-1}{2^x \ln(2)} \right]_{1}^{b} = \lim_{b \to \infty} \left( \frac{-1}{2^b \ln(2)} - \frac{-1}{2^1 \ln(2)} \right)$

As $b \to \infty$, the term $\frac{-1}{2^b \ln(2)}$ approaches $0$: $\int_{1}^{\infty} 2^{-x} dx =[...] 0 + \frac{1}{2\ln(2)} = \frac{1}{2\ln(2)}$ Since the integral converges to a finite number, the series converges by the Integral Test. We know from our previous work that this specific geometric series sums to 1.

Applying Integration by Parts, $\sum_{n=1}^\infty \frac{\ln(n)}{n^2}$

1. Step 1: Check Conditions. Let $f(x) = \frac{\ln(x)}{x^2}$. We must verify that it is positive, continuous, and decreasing for $x$ sufficiently large:
   Positive: For $x > 1$, $\ln(x) > 0$ and $x^2 > 0$, so $f(x)$ is positive.
   Continuous: The function is continuous for all $x > 0$.
   Decreasing: We find the derivative using the quotient rule: $f'(x) = \frac{x^2\left(\frac{1}{x}\right) - \ln(x)(2x)}{x^4} = \frac{x - 2x\ln(x)}{x^4} = \frac{1 - 2\ln(x)}{x^3}$
   For $f(x)$ to be decreasing, we need $f'(x) \leq 0$: $1 - 2\ln(x) \leq 0 \implies 2\ln(x) \geq 1 \implies \ln(x) \geq \frac{1}{2} \implies x \geq e^{1/2}$
   Since $e^{1/2} \approx 1.65$, the function is strictly decreasing on the interval $[2, \infty)$. This satisfies the requirement for the Integral Test.

2. Step 2: Evaluate the Improper Integral
   We can start the integral at $x=1$. Even though the function is only decreasing from $x=2$ onwards, the finite first term does not affect the overall convergence.
   $\int_{1}^{\infty} \frac{\ln(x)}{x^2} dx = \lim_{b \to \infty} \int_{1}^{b} \ln(x) \cdot x^{-2} dx$
   Use integration by parts: $u = \ln(x) \implies du = \frac{1}{x} dx$, $dv = x^{-2} dx \implies v = -\frac{1}{x}$
   $\int \frac{\ln(x)}{x^2} dx = -\frac{\ln(x)}{x} - \int \left(-\frac{1}{x}\right)\left(\frac{1}{x}\right) dx = -\frac{\ln(x)}{x} + \int x^{-2} dx = -\frac{\ln(x)}{x} - \frac{1}{x}$
   Now, evaluate the limits: $\lim_{b \to \infty} \left[ -\frac{\ln(x)}{x} - \frac{1}{x} \right]_{1}^{b} = \lim_{b \to \infty} \left( -\frac{\ln(b)}{b} - \frac{1}{b} \right) - \left( -\frac{\ln(1)}{1} - \frac{1}{1} \right)$
   To evaluate $\lim_{b \to \infty} \frac{\ln(b)}{b}$, apply L’Hôpital’s Rule ($\frac{\infty}{\infty}$ form): $\lim_{b \to \infty} \frac{\frac{1}{b}}{1} = 0$
   Therefore, the entire limit evaluates to: $(0 - 0) - (0 - 1) = 1$

3. Conclusion. Since the integral converges to 1, the series converges.

Recognizing the Arctangent Pattern, $\sum_{n=1}^\infty \frac{2}{n^2+6n+10}$

1. First, complete the square in the denominator: $n^2+6n+10 = (n^2+6n+9)+1 = (n+3)^2+1$.
   The series can be rewritten as: $\sum_{n=1}^\infty \frac{2}{(n+3)^2+1}$
2. Check Conditions. Let $f(x) = \frac{2}{(x+3)^2+1}$.
   Positive: The numerator is positive, and the denominator is always $\geq 1$.
   Continuous: The denominator $(x+3)^2+1$ can never be zero, so it is continuous everywhere.
   Decreasing: The denominator $(x+3)^2+1$ is an upward-opening parabola with a vertex at $x = -3$. For $x \geq 1$, the denominator is strictly increasing, which means the fraction as a whole is strictly decreasing.
3. Evaluate the Improper Integral
   $\int_{1}^{\infty} \frac{2}{(x+3)^2+1} dx = 2 \lim_{b \to \infty} \int_{1}^{b} \frac{1}{(x+3)^2+1} dx$
   Note: While trigonometric substitution works here, recognizing the standard integral form $\int \frac{1}{u^2+1}du = \arctan(u)$ is much faster.
   Let $u = x+3$, so $du = dx$: $2 \lim_{b \to \infty} \Big[ \arctan(x+3) \Big]_{1}^{b} = 2 \lim_{b \to \infty} \Big( \arctan(b+3) - \arctan(4) \Big)$.
   As $b \to \infty$, the arctangent approaches $\frac{\pi}{2}$: $\int_{1}^{\infty} \frac{2}{(x+3)^2+1} dx = 2 \left( \frac{\pi}{2} - \arctan(4) \right) = \pi - 2\arctan(4)$
4. Conclusion. Since the integral evaluates to a finite number, the series converges.

U-Substitution with Exponentials, $\sum_{n=1}^\infty \frac{e^{\frac{1}{n}}}{n^2}$

1. Check Conditions. Let $f(x) = \frac{e^{1/x}}{x^2}$.
   Positive: The exponential function is always positive, and $x^2 > 0$ for $x \geq 1$.
   Continuous: The only discontinuity is at $x=0$, which is outside our interval of $[1, \infty)$.
   Decreasing: Apply the quotient rule and chain rule to find $f'(x)$: $f'(x) = \frac{x^2 \left( e^{1/x} \cdot \frac{-1}{x^2} \right) - e^{1/x}(2x)}{x^4} = \frac{-e^{1/x} - 2xe^{1/x}}{x^4} = \frac{-e^{1/x}(1+2x)}{x^4}$
   Since $e^{1/x}$ is positive and $(1+2x)$ is positive for all $x \geq 1$, the numerator is negative while the denominator is positive. Therefore, $f'(x) \leq 0$, meaning the function is decreasing.

2. Evaluate the Improper Integral.
   $\int_{1}^{\infty} \frac{e^{1/x}}{x^2} dx$
   Use $u$-substitution. Let $u = \frac{1}{x}$, which means $du = -\frac{1}{x^2} dx$, or $-du = \frac{1}{x^2} dx$.
   We must also change the limits of integration. As $x \to \infty$, $u \to 0$. When $x = 1$, $u = 1$.
   $\int_{1}^{\infty} \frac{e^{1/x}}{x^2} dx = \int_{1}^{0} -e^u du = \int_{0}^{1} e^u du = \Big[ e^u \Big]_{0}^{1} = e^1 - e^0 = e - 1$

3. Conclusion. Since the integral converges to $e - 1$, the series converges.

Bibliography

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