If $\rho_{n+1}\leq\rho_{n} +\sigma_n,\;\;\forall\;\;n\geq N_0,$ then prove that $\lim\rho_n=0.$

Question

Good day, all!

Suppose $\{\rho_n\}$ and $\{\sigma_n\}$ are two sequences of non-negative real numbers such that for some real number $N_0\geq 1,$ the following recursion inequality holds:

$$\rho_{n+1}\leq\rho_{n} +\sigma_n,\;\;\forall\;\;n\geq N_0.$$

Prove that,

1. if $\sum_{n=1}^{\infty}\sigma_n<\infty,$ then $\lim\rho_n$ exists.
2. If $\sum_{n=1}^{\infty}\sigma_n<\infty,$ and $\rho_n$ has a subsequence converging to zero, then $\lim\rho_n=0.$

The first question was asked in my exam but I skipped it. I am stating the truth! Hence, I'll like to get it solved before the final exam. So please, can anyone help me? Thanks in advance.

Answer 1

Observe that once we prove assertion $1$, then the claim of $2$ follows readily. This is due to the fact, that if $\rho_n$ is convergent, than all its subsequences must be convergent, all with the same limit. Now, if we prove claim N$1$ and assume in addition that a subsequence of $\rho_n$ converges to $0$, then $\rho_n$, having the same limit as its subsequence, must converge to $0$.

We now prove claim $1$. Observe that for any $k\in \mathbb{N}$ and any $n\geq N_0$ we have $$ (1) \qquad \rho_{n+k} - \rho_n = \rho_{ n + k } - \rho_{ n + k -1} + \rho_{ n + k - 1 } - \rho_{ n + k -2} +...+\rho_{n+1} - \rho_n \leq \\ \sigma_{n+k-1} +...+\sigma_n, $$ where we apply the given inequality on each term $\rho_{n+k -i} - \rho_{n+k-i-1}$, for $i=0,....,k-1$.

From the above inequality, and the fact that $\rho_n \geq 0$, we have that $\rho_n$ is a bounded sequence. Now assume, for contradiction, that it does not converge. Hence, $\rho_n$, converges to different limits along two different subseqeunces. Namely there exists $n_{k}$ and $m_k$, and non negative numbers $a_1$ and $a_2$ such that $$ \rho_{n_k} \to a_1 \qquad \text{ and } \qquad \rho_{m_k} \to a_2, $$ where $a_1 \neq a_2$. From $(1)$, let us fix $n\geq N_0$, and choose $k$ such that $n+k = n_k$. Then, taking limit with respect to $k$ and using the fact that the series $\sigma_n$ converges, we get $$ (2) \qquad a_1 - \rho_n \leq \sum_{i = n}^\infty \sigma_i. $$ Now take $n\to \infty$ along the second subsequence $m_k$, we obtain $a_1 - a_2 \leq 0$, since convergence of $\sum_i \sigma_i$ implies that the right hand side of $(2)$ converges to $0$.

Since the process is symmetric (we could have started with $m_k$ and not $n_k$) we obtain also that $a_2 \leq a_1$ and hence they must be equal.

This contradicts our assumption that $a_1 \neq a_2 $ and hence the limit of $\rho_n$ exists.

Answer 2

Put $T_n=\sum_{k=1}^n \sigma_k$. Then $T_n$ is a convergent sequence. Now put $u_n=T_{n-1}-\rho_n$. From the inequality given, we see that $u_n$ is increasing for $n$ large, and of course (as $\rho_n \geq 0$) $\leq T_{n-1}$, hence bounded, hence convergent. It is easy to finish.

Answer 3

From $\rho_{n+1}-\rho_n\leq \sigma_n $ we obtain $ \rho_{n}- \rho_m \leq \sum_{i=m}^{n-1}\sigma _i \Longrightarrow \rho_{n} \leq \rho_m+\sum_{i=m}^{n-1}\sigma _i $.

Taking $\limsup$ in respect of $n$ we obtain $\limsup \rho _n \leq \sum_{i=m}^\infty \sigma _i +\rho _m$.

Taking now $\liminf$, we obtain $$ \limsup \rho _n \leq \liminf\rho _m $$ So, $\rho_n\longrightarrow x$ where $x \in [-\infty,\infty]$. Since $\rho _n$ is non-negative and upper bounded by $\rho_0 +\sum_{i=1}^\infty \sigma _i$, we conclude that $x\in \mathbb{R}$.