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Question: Let $\{a_n\}$ and $\{b_n\}$ be sequences such that:

$$b_{n+1} = a_n + a_{n+1}$$

for every positive integer $n$, and suppose that $\{b_n\}$ is convergent. Prove that $\lim_{n\to\infty} \left(\frac{a_n}{n}\right)=0$.

Things I know:

  • Since $\{b_n\}$ is convergent with limit $L$ then : $$ \forall\epsilon>0, \exists N\in\mathbb{N}:\forall n\ge N, |b_n-L|<\epsilon $$
  • $b_{n+1}=a_n+a_{n+1} \implies a_n = b_{n+1}-a_{n+1}$
  • $\lim_{n\to\infty} \left(\frac{a_n}{n}\right)=0 \implies \lim_{n\to\infty} \left(\frac{ b_{n+1}-a_{n+1}}{n}\right) = 0$

I'm not necessarily sure how to go about proving this using the above information. It's clear that I have to show that the difference between $\{b_n\}$ and $\{a_n\}$ converges to a point, and this might have something to do with knowing that a convergent sequences is always Cauchy.

Any help or direction would be appreciated.

Thank you

Edit: I would like to use an approach that involves the following topics: Convergent Sequences, Cauchy Sequences, Supremums and Infimums, Limits, Epsilon Characterizations, Bounded Sequences, Bolzano-Weierstrass Theorem.

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First we do the case that $(b_n)$ converges to $0$, the general case will follow from this. Define $(c_n)$ by $c_1=a_1$ and $c_n=b_n$ for $n \geq 2$. Then I claim we have $$a_n=(-1)^n\sum_{k=1}^n (-1)^kc_k,$$ for all $n$. This can be seen by induction, it clearly holds for $n=1$ and if it holds for some $n \geq 1$ then $$a_{n+1}=c_{n+1}-a_n=c_{n+1}-(-1)^n\sum_{k=1}^n (-1)^kc_k=\sum_{k=1}^{n+1}(-1)^{n+1+k}c_k=(-1)^{n+1}\sum_{k=1}^{n+1} (-1)^kc_k,$$ so it holds for $n+1$. Now that we have this, we need to show $\frac{a_n}{n}$ goes to zero, which is equivalent to showing $|\frac{a_n}{n}|$ goes to zero. Notice the following $$|\frac{a_n}{n}| \leq \frac{1}{n}\sum_{k=1}^n |c_k|,$$ so we see that $|\frac{a_n}{n}| $ is bounded above by the average of the first $n$ terms of a sequence converging to $0$ since $c_k \rightarrow 0$. The average of the first $n$ elements of a sequence that converges to $0$, converges to $0$ (see the lemma beneith), that is we have $$\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n |c_k| = 0,$$ which establishes the result for the case that $(b_n)$ converges to $0$. If $(b_n)$ does not converge to zero but to some $L \in \mathbb{R}$, then the sequence $(b'_n)$ defined by $b'_n=b_n - L$ does converge to $0$ and if we define $a'_n=a_n-\frac{L}{2}$, we have $$a'_{n+1}+a'_{n}=b'_{n+1}$$ so by the preceding argument we have that $\lim \frac{a'_n}{n}=0$. But from this we have $$\lim \frac{a_n}{n}=\lim (\frac{a'_n}{n}+\frac{L/2}{n})=0,$$ which completes the proof.

EDIT: Now for the lemma:

Let $(s_n)$ be a sequence of real numbers converging to $0$. Then the sequence $(t_n)$ defined by $$t_n=\frac{1}{n}\sum_{k=1}^n s_k$$ converges to $0$ as well.

Proof Let $\epsilon>0$. There is some $N_1 \in \mathbb{N}$ such that if $n>N_1$ then $|s_n|<\frac{\epsilon}{2}$. We can now find $N_2>N_1$ such that if $n>N_2$, then $$|\frac{\sum_{k=1}^{N_1} s_k}{n}|<\frac{\epsilon}{2}.$$ Now by the triangle inequality, if $n>N_2$ we also have $$|t_n|<\frac{\sum_{k=1}^{N_1} |s_k| + \sum_{k=N_1+1}^n |s_k|}{n} < \frac{\epsilon}{2}+\frac{n-N_1}{n} \frac{\epsilon}{2}<\epsilon.$$

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  • $\begingroup$ Firstly, thank you for the detailed answer. Although I do understand your method and approach, the book does not touch series or summations until after this chapter, and so I don't know how I would arrive at your claim for the claim of $a_n$ in the form of a summation.Is there anyway to go about this only using things related to convergent sequences, limsups liminfs, and their properties? I would also appreciate a proof in regards to the average converging to 0. $\endgroup$ – Hushus46 Oct 13 '17 at 0:18
  • $\begingroup$ Sure, shall I edit it into my answer? $\endgroup$ – M. Van Oct 13 '17 at 7:48
  • $\begingroup$ That would be fine, thank you very much again! $\endgroup$ – Hushus46 Oct 13 '17 at 7:49
  • $\begingroup$ And to answer your questions: I arrived at the conclusion about $(a_n)$ when I started realizing that if I'd treat the $(b_n)$ as a "given" sequence and $(a_n)$ as a sequence inductively constructed by $a_{n+1}=b_{n+1}-a_n$, I could get a closed form for $(a_n)$ in terms of $(b_n)$ and $a_1$. (write out some terms and you can see it happening, $a_2=b_2-a_1$, $a_3=b_3-a_2=b_3-b_2+a_1$, etc). Moreover, I claim I only use things related to sequences and no theory about series! Maybe it is confusing since I use sigma notation but I could use other notation as well. $\endgroup$ – M. Van Oct 13 '17 at 8:49

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