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Find:

$(a) \lim\limits_{n\to \infty} \left(1+e-\displaystyle\sum_{i=0}^n \frac1{k!}\right)^{n!}$

$(b) \lim\limits_{n\to \infty} \left(1+e-\displaystyle\sum_{i=0}^n \frac1{k!}\right)^{(n+1)!}$

I know $\lim\limits_{n\to \infty} \displaystyle\sum_{i=0}^n \dfrac1{i!} = e$, but I don't know how to start the above :(

Thanks for (a). How to do (b)?

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Let

$$s_n = \sum_{i=0}^n \frac{1}{i!}$$

Since you already know $s_n \to e$, we can use

$$\lim_{x \to x_0 }(1+f(x))^{\frac{1}{f(x)}} = e,\ \text{when}\ \lim_{x\to x_0}f(x) = 0$$

to write the first limit (a) as:

$$\lim_{n\to \infty}(1+e-s_n)^{n!} = \lim_{n\to \infty}\left[(1+e-s_n)^{\frac{1}{e-s_n}}\right]^{n! (e-s_n)} = e^{L}$$

where

$$L = \lim_{n\to \infty} n!(e-s_n)$$

We can evaluate $L$ using Cesaro-Stolz:

$$L = \lim_{n\to \infty} \frac{e-s_n}{\frac{!}{n!}} = \frac{s_n-s_{n+1}}{\frac{1}{(n+1)!}-\frac{1}{n!}} = \lim_{n\to \infty}\frac{-\frac{1}{(n+1)!}}{-\frac{n}{(n+1)!}}=\lim_{n\to \infty}\frac{1}{n} = 0$$

So the first limit equals $e^0 = 1$. We can solve (b) very similarly, I leave it to you (the result is $e$).

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  • $\begingroup$ you wrote $\lim_{f(x) \to 0}(1+f(x))^{\frac{1}{f(x)}} = e$, but I think you meant "x" instead of $f(x)$ right? $\endgroup$ – Olivier Roche Feb 2 at 19:43
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    $\begingroup$ @OlivierRoche, I meant this limit is true for any $f(x)$ whose limit is $0$. Maybe the formulation is not clear enough. I edited. $\endgroup$ – LHF Feb 2 at 19:45
  • $\begingroup$ All clear, thanks. But then I get $L = \lim \frac{n!}{e-s_n}$. What did I miss? $\endgroup$ – Olivier Roche Feb 2 at 20:07
  • $\begingroup$ @OlivierRoche, In the exponent you write $n! = \frac{1}{e-s_n} \cdot n!\cdot (e-s_n)$. But $\frac{1}{e-s_n}$ vanishes with the limit I wrote (to get that $e$) and in the exponent remains $n!\cdot (e-s_n)$ $\endgroup$ – LHF Feb 2 at 20:11

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