📂Calculus

Eigenvalue Criterion

Summary¹

For the series Series $\sum\limits_{n=0}^{\infty} a_{n}$, let $\lim\limits_{n \to \infty} \sqrt[n]{\left| a_{n} \right|} = L$ be given.

(a) If $L < 1$, the series absolutely converges.

(b) If $L > 1$ or $L = \infty$, the series diverges.

(c) If $L = 1$, it cannot be determined.

Explanation

If $L = 1$, it cannot be determined, so other methods must be used to judge whether the series converges or diverges. The statement of the theorem is similar to Non-Decision Method.

(c)

If $L = 1$, it could either converge or diverge. Looking at the series $\sum \dfrac{1}{n^{2}}$, $$ \lim\limits_{n \to \infty} \sqrt[n]{\left| \dfrac{1}{n^{2}} \right|} = \lim\limits_{n \to \infty} n^{-2/n} = \lim\limits_{n \to \infty} (e^{\ln (n^{-2/n})}) = \lim\limits_{n \to \infty} (e^{ (-2/n) \ln (n)}) = e^{\lim\limits_{n \to \infty} (-2/n) \ln n} $$

The last equality holds because the exponential function is a continuous function. By L’Hopital’s Rule, $$ \lim\limits_{n \to \infty} (-2/n) \ln n = -2 \lim\limits_{n \to \infty} \dfrac{\ln n}{n} = -2 \lim\limits_{n \to \infty} \dfrac{1}{n} = 0 $$ Thus, $L = 1$. The series converges because it is the $p-$ series, which satisfies $p=2$.

On the other hand, considering the series $\sum \left( \dfrac{n+1}{n} \right)^{n}$, $$ \lim\limits_{n \to \infty} \sqrt[n]{ \left( \dfrac{n+1}{n} \right)^{n} } = \lim\limits_{n \to \infty} \dfrac{n+1}{n} = 1 $$

Thus, it is $L = 1$, but since it satisfies $\lim\limits_{n \to \infty} \left( \dfrac{n+1}{n} \right)^{n} = \lim\limits_{n \to \infty} \left( 1 + \dfrac{1}{n} \right)^{n} = e$ by the divergence test, the series diverges. These two examples show that if $L = 1$, one cannot determine the convergence of the series.

Proof

(a)

The idea of the proof is to compare it with a converging geometric series. Since $L \lt 1$, there exists a positive number $r$ that satisfies $L \lt r \lt 1$. Then, from $\lim\limits_{n \to \infty} \sqrt[n]{\left| a_{n} \right|} = L$, the following holds for some sufficiently large $N$.

$$ \sqrt[n]{\left| a_{n} \right|} \lt r, \quad \text{for all } n \ge N $$

Rewriting it,

$$ \left| a_{n} \right| \lt r^{n}, \quad \text{for all } n \ge N $$

However, the series $\sum\limits_{n = N}^{\infty} r^{n}$ is a geometric series and satisfies $|r| \lt 1$, so it converges. Therefore, by comparison test, $\sum\limits_{n = N}^{\infty} |a_{n}|$ also converges. In other words, $\sum a_{n}$ absolutely converges. (Finite terms before large $N$ do not affect the convergence of the series)

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(b)

If $L > 1$ or $L = \infty$, then the following holds for sufficiently large $N$.

$$ \sqrt[n]{\left| a_{n} \right|} \gt 1, \quad \text{for all } n \ge N $$

$$ \implies \left| a_{n} \right| \gt 1, \quad \text{for all } n \ge N $$

Therefore, $\lim\limits_{n \to \infty} a_{n} \ne 0$, and by the divergence test, $\sum a_{n}$ diverges.

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