$ u_k \leq \alpha \exp(kh\beta) $ (proof by induction ) Let $ (u_k)_{k={0,...,K}}  \subset \mathbb{R} $ be a non-negative sequence and let $ \alpha $ , $ \beta \in \mathbb{R} $ with $ \beta \geq 0 $ and $ h > 0$, so that
$$ u_k = \alpha + h\beta\sum_{l=0}^{k-1} u_l \qquad \forall k = 0,...,K$$
Show: for all $ k = 0,...,K $ the following inequality holds:
$$ u_k \leq \alpha \exp(kh\beta) $$
My attempt:
As you can read in the title I wanted to show this inequality by induction. So I have the base case:
$k=0$ :
$ u_0 = \alpha + h\beta\sum_{l=0}^{0-1} u_l = \alpha = \alpha \exp(0h\beta) $
Ok now I assume the statement is true for a $ k = 0,...,K $.
Now the inductive step:
We want:
$u_{k+1} \leq \alpha \exp((k+1)h\beta)  $
I started with the definition of $ u_{k+1} $:
$ u_{k+1} = \alpha + h\beta\sum_{l=0}^{k} u_l  = \alpha + h\beta\sum_{l=0}^{k-1} u_l + h\beta u_k = u_k + h\beta u_k = u_k ( h \beta +1) $.
Im not sure if I'm right at this point, but I could use my indutive hypothesis.
$ u_k ( h \beta +1) \leq \alpha \exp(kh\beta)( h \beta +1)$.
But unfortunately thats not what I want. Maybe I did something wrong? Can you help me out ?
Thank you in advance.
 A: You didn't really do anything wrong, although a fairly minor point is that your use of variables $k$ and $K$ is somewhat inconsistent. To finish your work, first note you got
$$u_{k+1} = u_k(h\beta+1) \leq \alpha\exp(kh\beta)(h\beta+1) \tag{1}\label{eq1A}$$
The problem states $u_k$ is a non-negative sequence so, with $u_0 = \alpha$, then $\alpha \ge 0$. Since $\exp((k+1)h\beta) = \exp(kh\beta)\exp(h\beta)$, with
$$h\beta + 1 \le \exp(h\beta) \implies \alpha\exp(kh\beta)(h\beta+1) \le \alpha\exp((k+1)h\beta) \tag{2}\label{eq2A}$$
it's sufficient to just prove the LHS. Since $\beta \ge 0$ and $h \gt 0$, then $x = h\beta \ge 0$. Consider
$$f(x) = e^{x}-x-1 \tag{3}\label{eq3A}$$
Note $e^x = 1 + x + \frac{x^2}{2!} + \ldots \; \to \; f(x) = \frac{x^2}{2!} + \ldots$, then $f(x) \ge 0 \; \forall \; x \ge 0$. Alternatively, $f(0) = e^{0}-0-1 = 0$ and
$$f'(x) = e^{x} - 1 \ge 0 \; \; \forall \; x \ge 0 \tag{4}\label{eq4A}$$
In either case, $\forall \; x = h\beta \ge 0$, we get $f(x) \ge 0 \; \to \; x + 1 \le \exp(x) \; \to \; h\beta + 1 \le \exp(h\beta)$. This shows the LHS of \eqref{eq2A} is true, with this completing the induction step because \eqref{eq1A} and \eqref{eq2A} combined gives $u_{k+1} \le \alpha\exp((k+1)h\beta)$.
