Apply AM-GM on $k$ copies of $a_k^{k+1}$ and $1$ copy of $\frac1{2^{k+1}}$ to get \begin{align*} \frac12 a_k^k & \le \frac{ka_k^{k+1} + \frac1{2^{k+1}}}{k+1}\\ \implies (k+1) a_k^k & \le 2ka_k^{k+1} + \frac1{2^k}. \end{align*}Summing over all $k$, we get \begin{align*} \sum (k+1) a_k^k & \le 2\sum ka_k^{k+1} + \sum \frac1{2^k}\\ & < 2 \cdot 1 + 1\\ & = 3. \end{align*}