Series and concavity

Question

If $u(x)$ is strictly concave, can I say:

$$ \sum_{n=0}^{\infty} \left(\frac{1}{2}\right)^{n+1}\cdot u(n) < \infty. $$

I am having trouble finding counterexamples.

Thanks.

Answer 1

Because $u$ is strictly concave, for $n \ge 1$ we have $$u(n) \le (u(1) - u(0))n + u(0),$$ so that $$\sum_{n=0}^\infty 2^{-n-1} u(n) \le u(0) + \sum_{n=1}^\infty 2^{-n-1} ((u(1) - u(0))n + u(0)) < +\infty.$$