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According to a theorem in my book, it follows that

"If the series $\sum_{k=1}^{\infty}a_k$ converges $\Longrightarrow \lim_{k\rightarrow\infty}a_k=0$, but the reverse implication is not true."

This clearly means that I can't use this theorem to conclude whether a series converges.

Questions:

  1. What use is this theorem for then?
  2. Is this statement true: $$\lim_{k\rightarrow\infty}a_k=\pm\infty\Longrightarrow \sum_{k=1}^{\infty}a_k \ \text{diverges} \ ?$$
  3. What if $k$ starts from $0$, will above theorems still hold?
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  • $\begingroup$ It's a divergence criterion $\lnot(a_k \to 0) \implies \sum a_k$ diverges. $\endgroup$ Dec 30 '17 at 14:38
  • $\begingroup$ Note that in $p$-adic Analysis, a series is convergent if and only if its general term tends to $0$. $\endgroup$
    – Bernard
    Dec 30 '17 at 14:42
  • $\begingroup$ Yes, but I can't compute the limit of the general term to conclude if the series is convergent, according to that theorem above? $\endgroup$
    – Parseval
    Dec 30 '17 at 14:49
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1.The theorem is mainly used for establish the divergence of a series. (Contrapositive to the original).

2.Your statement is true. Actually, whenever the limit is not going to 0, you can conclude the series is divergent.

3.$k=0$ is fine. Actually, you can trim away the first finite terms in a series and its divergence could still be established with this theorem.

(Some more thoughts for students studying real analysis: We can trim off the first finite terms of a sequence due to this theorem:
Let $X=\{X_n:n\in\mathbb N\}$ be a sequence of real numbers and let $m\in \mathbb N$. Then the m-tail $X_m=\{X_{n+m}:n\in\mathbb N\}$ of X converges if and only if X converges. In this case, $\lim_{x \to \infty} X_m=\lim_{x \to \infty} X$ )

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  • $\begingroup$ One more question, what would happen if $\lim_{k\rightarrow\infty}a_k=m$? The number $m$ is just an arbitrary real constant. $\endgroup$
    – Parseval
    Dec 30 '17 at 15:01
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    $\begingroup$ As long as $m\neq 0$ we conclude that the series is divergent. $\endgroup$
    – Macrophage
    Dec 30 '17 at 15:04
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As for every implication, for negation we have that:

$$\lim_{k\rightarrow\infty}a_k\neq 0\Longrightarrow \sum_{k=1}^{\infty}a_k \ \text{does not converges}$$

It could diverge or oscillate.

This is used to prove easily that the series does not converges.

The series behaviuor is independent by a finite number of initial terms.

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In a nutshell:

$\lim _{k \rightarrow \infty }a_k = 0$ is

a necessary but not sufficient condition

for the conververgence of the series

$\sum_{k=0}^{\infty} a_k.$

Necessary :

Convergence of series implies

$\lim_{k \rightarrow \infty} a_k =0.$

The geometric series with terms: $a_k =(1/2)^k$ converges.

The criterion is not sufficient:

For the terms of the harmonic series: $a_k = 1/k $, we have

$\lim _{k \rightarrow \infty}a_k= 0.$

However: The series $\sum_{k=1}^{\infty} 1/k$ is not convergent.

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