📂Calculus

Non-deterministic Method

Theorem¹

For the series $\sum\limits_{n=0}^{\infty} a_{n}$, let $\lim\limits_{n \to \infty} \left| \dfrac{a_{n+1}}{a_{n}} \right| = L$.

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

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

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

Explanation

If $L = 1$, the convergence cannot be determined, and another method must be used to judge whether the series converges or diverges. The statement of the theorem is similar to the root test.

(c)

If $L = 1$, the series may either converge or diverge. For example, the two series below both satisfy $L = 1$, but the harmonic series on the left diverges, while the $p$-series on the right converges.

$$ \sum\limits_{n = 1}^{\infty} \dfrac{1}{n} \qquad \sum\limits_{n = 1}^{\infty} \dfrac{1}{n^{2}} $$

Proof

(a)

The idea of the proof is to compare with a convergent geometric series. Since $L \lt 1$, there exists a positive $r$ satisfying $L \lt r \lt 1$. Then, since $\lim\limits_{n \to \infty} | a_{n+1} / a_{n} | = L$, for sufficiently large $N$, we have the following.

$$ \left| \dfrac{a_{n+1}}{a_{n}} \right| \lt r, \quad \text{for all } n \ge N $$

Rewriting,

$$ \left| a_{n+1} \right| \lt \left| a_{n} \right|r, \quad \text{for all } n \ge N \tag{1} $$

Substituting $n = N$ into $(1)$,

$$ \left| a_{N+1} \right| \lt \left| a_{N} \right|r \tag{2} $$

Substituting $n = N+1$ into $(1)$, and by $(2)$, we obtain the following.

$$ \left| a_{N+2} \right| \lt \left| a_{N+1} \right|r \lt \left| a_{N} \right|r^{2} $$

In the same manner, we obtain the following formulas.

$$ \begin{align*} \left| a_{N+3} \right| &\lt \left| a_{N+2} \right|r \lt \left| a_{N+1}\right|r^{2} \lt \left| a_{N}\right|r^{3} \\ \left| a_{N+4} \right| &\lt \left| a_{N+3} \right|r \lt \left| a_{N+2}\right|r^{2} \lt \left| a_{N+1}\right|r^{3} \lt \left| a_{N}\right|r^{4} \\ &\vdots \\ \left| a_{N+k} \right| &\lt \left| a_{N}\right|r^{k},\quad \text{for all } k \ge 1 \end{align*} $$

Since the series $\sum\limits_{k = 1}^{\infty} |a_{N}|r^{k} = |a_{N}| r + |a_{N}| r^{2} + |a_{N}| r^{3} + \cdots $ is a geometric series and $|r| \lt 1$, it converges. Therefore, by the comparison test, $\sum\limits_{k=1}^{\infty} |a_{N+k}|$ also converges. That is, $\sum a_{n}$ absolutely converges. (The finite number of initial terms do not affect the convergence of the series)

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(b)

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

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

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

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

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