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If I know that for all $k$ it holds that $a_k \geq a_{k-1} \geq 0$, and $1 = \lim_{k\rightarrow \infty} a_k + a_{k-1}$. Is this sufficient to conclude that $\lim_{k\rightarrow \infty} a_k = \frac{1}{2}$?

I am not sure if it is allowed to split such that $\lim_{k\rightarrow \infty} a_k + a_{k-1} = \lim_{k\rightarrow \infty} a_k + \lim_{k\rightarrow \infty} a_{k-1} = 2 \lim_{k\rightarrow \infty} a_k$. I've looked up the axioms for limits on the internet, and found nothing that states that this is not permitted, yet it sounds weird to me.

Can I conclude $\lim_{k\rightarrow \infty} a_k = \frac{1}{2}$ just from this information?

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All you need to show is that the limit $\lim_{k\to\infty}a_k$ exists.

It is monotone by assumption. Show that it is bounded also using the other assumption.


To show boundedness of $\{a_k\}$, note that $a_k+a_{k-1}$ is a bounded sequence and $$ a_k+a_{k-1}\geq 2a_{k-1}. $$

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  • $\begingroup$ I've thought about the bounded-part a little. Because $a_k \geq a_{k-1} \geq 0$ I think $2a_k$ is an upper bound, which is always less than 2. $\endgroup$ – Jasper Jun 8 '16 at 14:42
  • $\begingroup$ Since the limit of $a_k+a_{k-1}$ exists, the sequence $b_k:=a_k+a_{k-1}$ must be bounded. Now you know that $a_k$ is monotone, then... $\endgroup$ – Jack Jun 8 '16 at 14:46
  • $\begingroup$ To make it a little clearer, given that the limit of $a_k$ exists, $\lim_{k\to\infty}a_k=\lim_{k\to\infty}a_{k+1}$ $\endgroup$ – Yon Teh Jun 8 '16 at 14:49
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The splitting of limit of a sum into a sum of limits require that each of the limits exists individually. In your problem setting, before you split the limit, you have to prove $\lim_k a_k$ exists first.

Alternatively, you can also show the result by squeeze theorem, since we have

$$ \frac{1}{2}(a_{k-1} + a_k) \leq a_k \leq \frac{1}{2}(a_k + a_{k+1}).$$

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Hint: An increasing sequence has a limit. By assumption, this limit must be a real number. Furthermore, any subsequence of a converging sequence converges to the same limit.

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Because $\{a_k\}$ is bounded above by 1/2 and increasing, the sequence $\{a_k\}$ has a limit. The bound of $\frac{1}{2}$ comes from a short contradiction. If any $a_k$ was greater than $\frac{1}{2}$, then the other given limit would not hold. Hence, you can decompose: $$1 = \lim_{k \rightarrow \infty}(a_k + a_{k-1}) = \lim_{k \rightarrow \infty}(a_k) + \lim_{k \rightarrow \infty}(a_{k-1}) = 2 \times \lim_{k \rightarrow \infty}(a_k)$$ Then you have that $\lim_{k\rightarrow \infty} (a_k) = \frac{1}{2}$.

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  • $\begingroup$ Fixed. Read too fast. $\endgroup$ – Nick Potter Jun 8 '16 at 14:51

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