Let an algorithm compute $e$ by $e=\sum\limits_{k=0}^\infty 1/k!$ (assume that $fl(k)=k$). The last term of this summation will satisfy $1/k! < \epsilon_{mach}$.
My attempt at solution:
Let $m$ be the last term in the discrete summation. Then
$fl(\sum\limits_{k=0}^\infty 1/k!)=\sum\limits_{k=0}^m fl(1/k!)(1+\epsilon_1)=(1+\epsilon_1)\sum\limits_{k=0}^m 1/k! (m(1+\epsilon_2))\le (1+2\epsilon_{mach})\sum\limits_{k=0}^m 1/k! + O(\epsilon_{mach}^2)$
Which means the algorithm is stable, but not backward stable (since we have $1+2\epsilon_{mach}$ instead of $1+\epsilon_{mach}$). I'd appreciate if someone could let me know if my solution is correct. I'm concerned that the last inequality only shows that the algorithm is at least stable, which may include backward stability.