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Does $\lim_{n\to\infty}a_n \neq 0 \rightarrow \sum_{n=1}^\infty a_n$ diverges hold if $a_n$ has also negative terms?

I can prove that if $a_n \geq 0$ then the implication holds, but I'm not sure about how to prove/ disprove the case when negative terms are allowed

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  • $\begingroup$ Try proving the contrapositive $\endgroup$ – Simon S Oct 20 '15 at 21:45
  • $\begingroup$ Yes, it's true if $a_n$ has negative (or even complex) terms. As @SimonS suggests, try the contrapositive. In particular, note that if the series converges, then the sequence of partial sums is a Cauchy sequence. $\endgroup$ – Bungo Oct 20 '15 at 21:46
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    $\begingroup$ Niebla, your most recent edit has made the existing answers irrelevant, so I have rolled it back. See this discussion on meta. For the version involving $\lim n a_n$ see this question. $\endgroup$ – Antonio Vargas Oct 21 '15 at 0:56
  • $\begingroup$ Also this question. $\endgroup$ – Antonio Vargas Oct 21 '15 at 1:05
  • $\begingroup$ But again, if an is not decreasing and strictly positive, does that hold? $\endgroup$ – Niebla Oct 21 '15 at 2:24
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Suppose the series converges to a limit $\ell$. Then $A_n = \sum_{k=1}^n a_k$ and $B_n = \sum_{k=1}^{n-1} a_k$ both converge to $\ell$. But then $$ a_n = A_n - B_n \xrightarrow[n\to\infty]{} \ell - \ell = 0. $$ (using usual operation on limits: sum, subtraction.)

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  • $\begingroup$ That's definitely simpler than going with the definition of Cauchy sequences like I did. +1 $\endgroup$ – Patrick Da Silva Oct 20 '15 at 21:53
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If $\sum_{n \ge 0} a_n$ is a convergent series, then $\lim_{n \to \infty} a_n = 0$. Your statement is the converse to this one. The proof of this is very simple : if the sequence $S_N = \sum_{n=0}^N a_n$ is convergent (that is, if the series converges), then $S_N$ is a Cauchy sequence, hence

$$ \forall \varepsilon > 0, \exists M \text{ such that } \forall N,N' \ge M, \quad |S_N - S_{N'}| < \varepsilon. $$ Picking $n = N' = N-1$ bigger than $m = M+1$, this shows $$ \forall \varepsilon > 0, \exists m \text{ such that } \forall n \ge m, \quad |a_n| < \varepsilon. $$ since $S_n - S_{n-1} = a_n$. In other words, $a_n \to 0$.

Hope that helps,

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