Suppose a differentiable $f:\mathbb{R}\to(0,∞)$ satisfying $$\alpha f(x+y)=f(x)f(y)\ (\alpha >f(1)).$$ Express the following sum in terms of $ \alpha$ and $f(1)$: $$S=\sum_{i=1}^∞ f(i).$$
Method 1: Geometric progression
Since $0$ is out of the range of $f(x)$ as defined above, I noted that $f(0)=\alpha$.
$\alpha f(2)=f(1)^2 =>f(2)=\dfrac{f(1)^2}{\alpha}$
$ \alpha f(3)=f(2)f(1)=\dfrac{f(1)^3}{\alpha} =>f(3)=\dfrac{f(1)^3}{\alpha^2}$
Similarly, it is easy to note that $f(i) = \dfrac{f(1)^{i}}{\alpha^{i-1}}$
So the above summation is simply a geometric progression which converges since $\dfrac{f(1)}{\alpha}<1$;
$S=\dfrac{\alpha f(1)}{\alpha - f(1)}$
Method 2: Partial Differentiation
$\alpha f'(x+y)=f'(x)f(y)$ (Differentiating wrt x keeping y constant)
Setting $x=0$, and $y=t$, where $t$ is a variable parameter,
$\alpha f'(t) = f'(0)f(t)$
$=>\alpha \ln f(t) = f'(0) t + C$ and since $f(0)=\alpha$,
$=>\ln f(x)=\dfrac{f'(0)}{\alpha}x + \dfrac{C}{\alpha}$
$=> f(x) = Ae^{f'(0)x/\alpha}$ where $A=e^{C/\alpha}$
Plugging $f(1)$ into this function, $A=f(1)e^{-f'(0)/\alpha}$
Hence, the final function is $f(x) = f(1)e^{f'(0)/\alpha(x-1)}$ This will only converge when $f'(0)<0$. How can I prove that using the given condition on $alpha$ and $f(1)$?
And secondly, if I sum up $f(i)$ from $1$ to $∞$ now, the sum becomes;
$S=f(1)[1+e^{f'(0)/\alpha}+e^{2f'(0)/\alpha}+...]$
$=>S=\dfrac{f(1)}{1-e^{f'(0)/\alpha}}$
How can I reconcile this answer with the first one?
EDIT: I corrected the issue with f(0) and f'(0), very very sorry about that.
Figured it out, full credit to Ms. Kiwi; differentiating the function yields -
$f'(x) = \dfrac {f(1)f'(0)}{\alpha}e^{f'(0)(x-1)/\alpha}$
$\implies f'(0)=\dfrac {f(1)f'(0)}{\alpha}e^{-f'(0)/\alpha}$ and since $f'(0)$ is not equal to $0$ otherwise the function would be a constant function, which is not possible since $\alpha > f(1)$ and the functional relation would imply $f(x)=\alpha$ for all $x$.
Hence, $e^{f'(0)/\alpha} = \dfrac {f(1)} {\alpha} <1 ...(i)$
$\implies f'(0)<0$ (Taking natural log on both sides of the above equality)
which ensures that the sum always converges. Using $(i)$, we may also rewrite the second sum as;
$S=\dfrac {f(1)}{1-e^{f'(0)/\alpha}}$
$\implies S=\dfrac {f(1)}{1-f(1)/\alpha}$
$\implies S=\dfrac {\alpha f(1)}{\alpha -f(1)}$