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Given a convergent sequence $\{ a_n \}_{n=1}^{\infty}$ and a bounded sequence $\{ b_n \}_{n=1}^{\infty}$ such that $b_{n+1}\leq b_n+(a_{n+1}-a_n)$ for every $n\in\mathbb{N}.$ Prove $\{ b_n \}_{n=1}^{\infty}$ converges.

My try:

$\{ b_n \}_{n=1}^{\infty}$ is bounded $\Rightarrow \exists M>0\ \forall n\in\mathbb{N}:|b_n|\leq M $.

Re-arranging: $-M-(-M)\leq b_{n+1}-b_n\leq a_{n+1}-a_n$

And by squeeze theorem, we get $\lim _{n\to \infty }\left(b_{n+1}-b_n\right)=0$.

Am I going in the right direction? If so, how can I continue from here?

Any help appreciated.

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    The inequality $0\le b_{n+1}-b_n$ is false : you only have $-2M\le b_{n+1}-b_n$ – 2017-02-26
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    Try to show that $\{b_n - a_n\}_{n=1}^\infty$ converges. You can then say that the sum of $\{b_n - a_n\}$ and $\{a_n\}$ converges. – 2017-02-26
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    @TrevorNorton Can I do it like that $L-\epsilon\ < a_n\ < L+\epsilon\ \Rightarrow L-\epsilon-b_n\ < a_n-b_n\ < L+\epsilon-b_n\ \Rightarrow M+L-\epsilon\ < a_n-b_n\ – 2017-02-26
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    @Itay4 Yes, although I did not approach it from the $\epsilon$ definition of the limit. If we can show that $\{b_n-a_n\}$ is bounded and monotonic, we know it converges. Given what we know, this is a fairly straight-forward calculation. I will give the details in an answer below. – 2017-02-26

2 Answers 2

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Hint: Look at the sequence $\{b_n-a_n\}_{n=1}^\infty$.

From our initial assumption, we have that $b_{n+1}\leq b_n+(a_{n+1}-a_n)$ for all $n\in\mathbb N$. Rearranging gives $$b_{n+1} - a_{n+1} \leq b_n - a_n.$$ Thus the sequence $\{b_n - a_n\}$ is decreasing. Also, let $\{b_n\}$ and $\{a_n\}$ be bounded by $M_1$ and $M_2$ respectively. Then $$|b_n-a_n| \leq |b_n| +|a_n| \leq M_1 + M_2$$ for all $n\in\mathbb N$, i.e., $\{b_n - a_n\}$ is bounded. Therefore $\{b_n -a_n\}$ converges. Since the sum of convergent sequences converges, $\{(b_n-a_n) +a_n\} = \{b_n\}$ converges.

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Rearranging we get $b_{n+1}-b_{n}m$.

Since $\{b_n\}$ is bounded, we may define $l=\liminf_n b_n$ and we know that there is some convergent subsequence $b_{n_m}\to l$. Let $\epsilon>0$, since $\{a_n\}$ converges, we may take $N_1$ large enough so that $|a_n-a_m|<\epsilon/2$ for all $n,m>N_1$.

Take $m>N_1$ large enough so that $|b_{n_{m}}-l|<\epsilon/2$ and note that $n_{m}\geq m>N_1$. Hence, for $n> n_{m}$ we have $$b_n-b_{n_{m}}

We directly conclude that $\limsup_n b_n\leq l+\epsilon$ and we may even remove the $\epsilon$ since it was arbitrary. Hence, we obtain $$l=\liminf_n b_n\leq \limsup_n b_n\leq l,$$ giving the desired result.