limit and convergence rate of $\lim_{t\to\infty}\left(1-\frac{\log(ct + c + 1)}{t} - \frac{\log(c+1)}{t}\right)^{t}$

Question

I am trying to compute the limit and the rate of convergence of

$$\lim_{t\to\infty}\left(1-\frac{\log(ct + c + 1)}{t} - \frac{\log(c+1)}{t}\right)^{t}$$

where $c$ is a positive constant and $t \in \mathbb{N}$. How do you compute that?

Answer 1

Assuming $\log=\ln$, first get a respresentation that's a bit simpler using $a:=c(c+1)$ and $b:=(c+1)^2$:

$$\begin{align} f(t) := \Big(1-\frac{\ln(ct + c + 1)}{t} - \frac{\ln(c+1)}{t}\Big)^t &= \Big(1-\frac1t \ln((ct+c+1)(c+1))\Big)^t \\ &= \Big(1-\underbrace{\frac1t \ln(at+b)}_{\displaystyle =:x(t)}\Big)^t \\ \end{align}$$

To analyze this further, notice that $x(t)\to 0$ as $t\to\infty$ and assuming $a > 0$. This means we can take $\ln(f(t))$ and expand it according to

$$\ln(1-x) = -\sum_{k=1}^\infty x^k/k = -x+\Theta(x^2) \tag 1$$ because we'll have $|x(t)| < 1$ so that it's legitimate to apply that expansion.

$$\begin{align} \ln f(t) = \ln \big(1-x\big)^t &= t\ln (1-x) \\ &\stackrel{(1)}= -t \big( x + \Theta(x^2) \big)\\ &= -t \Big(\frac{\ln(at+b)}{t} + \Theta\Big(\frac{\ln^2(at+b)}{t^2}\Big) \Big) \\ &= - \ln(at+b) + \Theta\Big(\frac{\ln^2(at+b)}{t}\Big) \\ \end{align}$$

Finally, apply $\exp$ on both sides to undo the outer $\ln$ again: $$\begin{align} f(t) &= \frac1{at+b} \cdot \exp\Big(\Theta\Big(\frac{\ln^2(at+b)}{t}\Big)\Big) \\ &= \frac1{at+b} \cdot \Big(1+\Theta\Big(\frac{\ln^2(at+b)}{t}\Big)\Big)\tag{*}\\ &= \frac1{at+b} \cdot \Big(1+\Theta\Big(\frac{\ln^2t}{t}\Big)\Big) \\ \end{align}$$ where for the line $(*)$ we used that the argument of $\exp$ is small and thus $\exp y = 1 +\Theta(y)$. This means that $$\lim_{t\to\infty} f(t) = 0$$ and you can read off the speed of convergence resp. the error term.