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Jun 23, 2020 at 21:46 comment added Christian Remling This should have the same proof as the scalar calculus identity that you mention since all the operators involved here ($1,B$) commute, so the fact that we're dealing with operators rather than numbers can't really make itself felt.
Jun 23, 2020 at 20:32 comment added Nate Eldredge Oh you're right, never mind.
Jun 23, 2020 at 20:30 comment added MightyPower @Nate Eldredge, Why? I compare this with a similar limit for functions and $\lim_\limits{n \to \infty} \left(1 - \frac{a}{n} \right)^{- (n-p)} = e^a$, not $e^{pa}$.
Jun 23, 2020 at 20:27 comment added MightyPower @Nate Eldredge, Yes E - identity operator.
Jun 23, 2020 at 20:10 comment added Nate Eldredge Is $E$ the identity operator?
Jun 23, 2020 at 20:00 history asked MightyPower CC BY-SA 4.0