The Ratio and Root Tests for Series Convergence

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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.

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.

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 $\sum a_n$ and $\sum b_n$ be series with positive terms such that $0 \leq a_n \leq b_n$ for all $n \geq N$.

1. If the “bigger” series $\sum b_n$ converges, then the “smaller” series $\sum a_n$ converges.
2. If the “smaller” series $\sum a_n$ diverges, then the “bigger” series $\sum b_n$ diverges.
3. If the bigger series diverges, or the smaller series converges, the test is inconclusive.

The Ratio Test

The Ratio Test is particularly useful for series involving factorials (like $n!$) or exponential terms (like $a^n$). It measures the rate at which terms in the series grow or shrink relative to one another.

Theorem. The Ratio Test. Suppose we have the series $\sum_{n=1}^\infty a_n$, and let: $L = \lim_{n \to \infty} \left\vert{} \frac{a_{n+1}}{a_n} \right\vert{}$. Then, the following statements hold true:

1. If $L < 1$, the series $\sum_{n=1}^\infty a_n$ is absolutely convergentT (and therefore convergent).
2. If $L > 1$ (or if $L = \infty$), the series $\sum_{n=1}^\infty a_n$ is divergent.
3. If $L = 1$, the test is inconclusive. The series may diverge, converge conditionally, or converge absolutely. You must use a different test.

The Root Test

The Root Test is often used when the entire $n$-th term of a series is raised to the power of $n$.

Theorem. The Root Test. Suppose we have the series $\sum_{n=1}^\infty a_n$, and let: $L = \lim_{n \to \infty} \sqrt[n]{\vert{}a_n\vert{}} = \lim_{n \to \infty} \vert{}a_n\vert{}^{\frac{1}{n}}$. Then:

1. If $L < 1$, the series is absolutely convergent (and therefore convergent).
2. If $L > 1$ (or if $L = \infty$), the series is divergent.
3. If $L = 1$, the test is inconclusive. The series may diverge, converge conditionally, or converge absolutely.

Proof.

1. Part 1: If $L < 1$, the series converges.
   Assume $L < 1$. Because $L$ is strictly less than 1, there exists some real number $r$ such that $L < r < 1$.
   Since $\lim_{n \to \infty} \sqrt[n]{\vert{}a_n\vert{}} = L$, by the definition of a limit, the terms $\sqrt[n]{\vert{}a_n\vert{}}$ must eventually get arbitrarily close to $L$. Therefore, there must exist some integer $N$ such that for all $n \ge N$: $\sqrt[n]{\vert{}a_n\vert{}} < r$
   Raising both sides to the $n$-th power yields: $\vert{}a_n\vert{} < r^n$
   Now, consider the geometric series $\sum r^n$. Because $0 < r < 1$, this geometric series converges.
   Since $\vert{}a_n\vert{} < r^n$ for all $n \ge N$, the series $\sum_{n=N}^\infty \vert{}a_n\vert{}$ converges by the Direct Comparison Test.
   Because adding a finite number of initial terms back to the sum does not change its convergence, the entire series $\sum_{n=1}^\infty \vert{}a_n\vert{}$ converges absolutely.

2. Part 2: If $L > 1$, the series diverges.
   Assume $L > 1$. By the definition of a limit, since the terms $\sqrt[n]{\vert{}a_n\vert{}}$ are approaching $L$, they must eventually become greater than 1.
   There exists an integer $N$ such that for all $n \ge N$: $\sqrt[n]{\vert{}a_n\vert{}} > 1$
   Raising both sides to the $n$-th power yields: $\vert{}a_n\vert{} > 1^n = 1$
   Because the absolute value of the terms is strictly greater than 1 for all $n \ge N$, the limit of the sequence $a_n$ as $n \to \infty$ cannot possibly be $0$.
   By the Divergence Test (also known as the $n$-th Term Test), since $\lim_{n \to \infty} a_n \neq 0$, the series $\sum a_n$ diverges.

Part 3: If $L = 1$, the test is inconclusive.
To prove this, we simply look at examples where $L=1$ yields different results:
$\star$ $\sum_{n=1}^\infty \frac{1}{n}$ (Harmonic series): $L = 1$, and it diverges.
$\sqrt[n]{\left|\frac{1}{n}\right|} = \left(\frac{1}{n}\right)^{1/n} = n^{-1/n}.$
$\lim_{n\to\infty} n^{1/n} = 1 \implies \lim_{n\to\infty} n^{-1/n} = \frac{1}{\lim_{n\to\infty} n^{1/n}} = \frac{1}{1} = 1$
$\star$ $\sum_{n=1}^\infty \frac{1}{n^2}$ ($p$-series): $L = 1$, and it converges absolutely.
$\star$ $\sum_{n=1}^\infty \frac{(-1)^n}{n}$ (Alternating Harmonic): $L = 1$, and it converges conditionally.

Solved examples

• Simple Ratio Test. Determine convergence or divergence of the series $\sum_{n=1}^\infty \frac{n}{5^n}$.

1. Apply the Ratio Test: $L = \lim_{n \to \infty} \left\vert{} \frac{\frac{n+1}{5^{n+1}}}{\frac{n}{5^n}} \right\vert{} = \lim_{n \to \infty} \left( \frac{n+1}{5^{n+1}} \cdot \frac{5^n}{n} \right) = \lim_{n \to \infty} \left( \frac{n+1}{n} \cdot \frac{5^n}{5^{n+1}} \right) = \lim_{n \to \infty} \left( \frac{n+1}{n} \cdot \frac{1}{5} \right) = \frac{1}{5}$ \
2. Since $L = \frac{1}{5} < 1$, the series converges.

• Radius and Interval of Convergence. Find the radius and interval of convergence of the series $\sum_{n=1}^\infty \frac{x^n}{n}$.

1. Apply the Ratio Test: $L = \lim_{n \to \infty} \left\vert{} \frac{\frac{x^{n+1}}{n+1}}{\frac{x^n}{n}} \right\vert{} = \lim_{n \to \infty} \left\vert{} \frac{x^{n+1}}{n+1} \cdot \frac{n}{x^n} \right\vert{} = \lim_{n \to \infty} \vert{}x\vert{} \frac{n}{n+1} = \vert{}x\vert{} \cdot 1 = \vert{}x\vert{}$
2. For the series to converge, we need $L < 1 \implies \vert{}x\vert{} < 1$. The radius of convergence is $R = 1$.
3. Now we check the endpoints $x = \pm 1$:
   If $x = 1$, the series becomes $\sum_{n=1}^\infty \frac{1}{n}$, which is the Harmonic Series and diverges.
   If $x = -1$, the series becomes $\sum_{n=1}^\infty \frac{(-1)^n}{n}$, which is the Alternating Harmonic Series and converges (by the Alternating Series Test). \
4. The interval of convergence is $I = [-1, 1)$ and we may think of $\sum_{n=1}^\infty \frac{x^n}{n}$ as defining a function $f(x) = \sum_{n=1}^\infty \frac{x^n}{n}$ from the interval of convergence to the real numbers $\mathbb{R}$.

• Factorials and the Ratio Test. Find the radius and interval of convergence of the series $\sum_{n=0}^\infty (-1)^n \frac{x^{2n}}{(2n)!}$.

1. Apply the Ratio Test: $L = \lim_{n \to \infty} \left\vert{} \frac{(-1)^{n+1} \frac{x^{2(n+1)}}{(2(n+1))!}}{(-1)^n \frac{x^{2n}}{(2n)!}} \right\vert{} = \lim_{n \to \infty} \left\vert{} \frac{x^{2n+2}}{(2n+2)!} \cdot \frac{(2n)!}{x^{2n}} \right\vert{}$
   Expand the factorial: $(2n+2)! = (2n+2)(2n+1)(2n)!$, so we get: $= \lim_{n \to \infty} \left\vert{} \frac{x^2}{(2n+2)(2n+1)} \right\vert{} = x^2 \lim_{n \to \infty} \frac{1}{(2n+2)(2n+1)} = 0$
2. Because $L = 0 < 1$ regardless of the value of $x$, the series always converges.
3. $R = \infty$, and $I = (-\infty, \infty)$.

• A Shifted Power Series. Find the radius and interval of convergence of the series $\sum_{n=0}^\infty \frac{(x-3)^n}{n^7+1}$.

1. Apply the Ratio Test: $L = \lim_{n \to \infty} \left\vert{} \frac{\frac{(x-3)^{n+1}}{(n+1)^7+1}}{\frac{(x-3)^n}{n^7+1}} \right\vert{} = \lim_{n \to \infty} \left\vert{} (x-3) \frac{n^7+1}{(n+1)^7+1} \right\vert{}$
   Divide numerator and denominator by $n^7$: $L = \vert{}x-3\vert{} \lim_{n \to \infty} \frac{1 + \frac{1}{n^7}}{\left(1+\frac{1}{n}\right)^7 + \frac{1}{n^7}} = \vert{}x-3\vert{} \cdot \frac{1+0}{1^7+0} = \vert{}x-3\vert{}$
   We require $\vert{}x-3\vert{} < 1$. The radius is $R = 1$. The endpoints are $3 - 1 = 2$ and $3 + 1 = 4$.
2. If $x = 2$: $\sum_{n=0}^\infty \frac{(-1)^n}{n^7+1}$. This converges by the Alternating Series Test.
3. If $x = 4$: $\sum_{n=0}^\infty \frac{1}{n^7+1}$. Compare this to $\sum \frac{1}{n^7}$. Since $\frac{1}{n^7+1} < \frac{1}{n^7}$, and $\sum \frac{1}{n^7}$ converges ($p$-series, $p=7>1$), our series converges by the Direct Comparison Test.
4. The interval of convergence is $I = [2, 4]$.

• Root Test with a Variable. Determine convergence or divergence of the series $\sum_{n=1}^\infty n^nx^n$.

1. Apply the Root Test: $L = \lim_{n \to \infty} \sqrt[n]{\vert{}n^nx^n\vert{}} = \lim_{n \to \infty} n\vert{}x\vert{} = \vert{}x\vert{} \lim_{n \to \infty} n$
2. Unless $x = 0$, this limit approaches $\infty$. Therefore, $L < 1$ only when $x = 0$.
3. $R = 0$ and $I = \{0\}$.

• Root Test on a Complex Limit. Determine convergence or divergence of the series $\sum_{n=1}^\infty \left(1+\frac{1}{n}\right)^{n^2}$.

1. The index starts at $n=1$ to avoid division by zero.
2. Apply the Root Test: $L = \lim_{n \to \infty} \sqrt[n]{\left(1+\frac{1}{n}\right)^{n^2}} = \lim_{n \to \infty} \left(1+\frac{1}{n}\right)^{\frac{n^2}{n}} = \lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n$
   This is the classic limit definition of Euler’s number $e$: $L = e \approx 2.718$
3. Conclusion. Since $L > 1$, the series diverges.

• Evaluating $\lim n^{1/n}$. Determine convergence or divergence of the series $\sum_{n=1}^\infty \frac{n}{2^n}$.

1. Apply the Root Test: $L = \lim_{n \to \infty} \sqrt[n]{\frac{n}{2^n}} = \lim_{n \to \infty} \frac{n^{\frac{1}{n}}}{2} = \frac{1}{2} \lim_{n \to \infty} n^{\frac{1}{n}}$
2. To evaluate $\lim_{n \to \infty} n^{\frac{1}{n}}$, let $y = n^{\frac{1}{n}}$. Then, $\ln(y) = \ln(n^{\frac{1}{n}}) = \frac{1}{n} \cdot \ln(n) = \frac{\ln(n)}{n}$.
   $\lim_{n \to \infty} \ln(y) = \lim_{n \to \infty} \frac{\ln(n)}{n}$
   This is an $[\infty/\infty]$ indeterminate form. We solve it by applying the L’Hôpital’s Rule: $\lim_{n \to \infty} \frac{\frac{1}{n}}{1} = 0$
   So, $\lim_{n \to \infty} \ln(y) = 0 \implies \lim_{n \to \infty} y = e^{\lim_{n \to \infty} \ln(y)} = e^0 = 1$.
3. Now, return to $L$: $L = \frac{1}{2} \lim_{n \to \infty} n^{\frac{1}{n}} = \frac{1}{2} \cdot 1 = \frac{1}{2}$
4. Since $L < 1$, the series converges.

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