How can I prove that $\{V_k\}$ is convergent? I tried to build a general term for $V_k$ but I don't get to anything.
Let $\{V_k\}$, $\{\lambda_k\}$, $\{B_k\}$ be non negative sequences of real numbers that satisfy
\begin{equation*} V_k \le (1+\lambda_k)V_{k-1} + B_k \end{equation*}
where $\sum\limits_{k=1}^\infty B_k < \infty$, $\sum\limits_{k=1}^\infty\lambda_k < \infty$.
Prove that $\{V_k\}$ is convergent.
Many thanks in advance
On
Hint: By induction, one can show \begin{align} V_k-V_{k-1} \leq&\ \lambda_k\prod^{k-1}_{i=2}(1+\lambda_i)V_1+\left(B_k+\lambda_kB_{k-1}+\ldots +\lambda_k\prod^{k-1}_{i=2}(1+\lambda_i)B_1\right)\\ \leq&\ \lambda_k \left\{e^{-\lambda}(V_1+\lambda B) \right\}+B_k. \end{align} where $\lambda = \sum \lambda_i$ and $B= \sum B_i$. Then we see that \begin{align} V_n-V_m=\sum^n_{k=m+1} (V_k-V_{k-1}) \leq e^{-\lambda}(V_1+\lambda B) \left(\sum^n_{k=m+1}\lambda_k\right) + \left(\sum^n_{k=m+1}B_k\right) \end{align} which shows that the sequence is Cauchy. Hence $\{V_k\}$ is convergent.
Hint. Note that for $n\geq m\geq 1$, $$V_n-V_m=\sum_{k=m+1}^n(V_k-V_{k-1}) \le \sum_{k=m+1}^n \lambda_kV_{k-1} + \sum_{k=m+1}^nB_k$$ Hence if we show that $(V_n)_n$ is bounded by some non-negative constant $M$ then $$|V_n-V_m|=|\sum_{k=m+1}^n(V_k-V_{k-1})| \le M\sum_{k=m+1}^n \lambda_k + \sum_{k=m+1}^nB_k$$ which implies that $(V_n)$ is a Cauchy sequence since $\sum\limits_{k=1}^\infty\lambda_k < \infty$ and $\sum\limits_{k=1}^\infty B_k < \infty$.