One more limit with sum of reciprocals of binomial coefficients

Question

It may be found interesting that the limit $$L=\lim_{n \rightarrow \infty} \left (\sum_{k=0}^N \frac{1}{{n \choose k}} \right)^n,$$ diverges when (i) $N=n$, converges to different values when (ii) $N=n-1$ and when (iii) $N=n-2$.Find $L$ in the last two cases.

Answer 1

$$\mbox{Let} ~~ S_N= \sum_{0}^N \frac{1}{n \choose k} ~ \mbox{then}~ L=\lim_{n\rightarrow \infty}(S_N)^n$$ Case (1): When $N=n$: $$ S_n=1+\frac{1}{n\choose 1}+\frac{1}{n\choose 2}+\frac{1}{n\choose 3}+....+\frac{1}{n\choose k}+...+\frac{1}{n\choose 2}+\frac{1}{n\choose 1}+1 \Rightarrow S_n > 2 \Rightarrow L= \lim_{n\rightarrow \infty} 2^n= \infty ~ \mbox{(divergent)}.$$ This divergence has also been pointed out by kccu.

Case (ii): When $N=n-1$ $$S_{n-1}=1+\frac{1}{n\choose 1}+\frac{1}{n\choose 2}+\frac{1}{n\choose 3}+....+\frac{1}{n\choose k}+...+\frac{1}{n\choose 2}+\frac{1}{n\choose 1} \Rightarrow$$ $$S_{n-2}=1+\frac{2}{n\choose 1}+\frac{2}{n\choose 2}+\frac{1}{n\choose 3}+\frac{1}{n\choose 4}+\frac{1}{n\choose 5}+...+\frac{1}{n\choose n-3}.$$ $$\Rightarrow 1+\frac{2}{n\choose 1} < S_{n-1}<1+\frac{2}{n\choose 1}+\frac{2}{n\choose 2}+(n-5)\frac{1}{n\choose 3},~~ \mbox{since}~~ {n\choose 3} \le {n \choose j}, \forall ~ j \le n-3. $$ $$\Rightarrow \lim_{n\rightarrow \infty} \left(1+\frac{2}{n\choose 1}\right)^n <\lim_{n \rightarrow \infty} (S_{n-1})^n < \lim_{n \rightarrow \infty}\left(1+\frac{2}{n\choose 1}+\frac{2}{n\choose 2}+(n-5)\frac{1}{n\choose 3}\right)^n.$$ $$\Rightarrow \lim_{n\rightarrow \infty} \left(1+\frac{2}{n}\right)^n <\lim_{n \rightarrow \infty} (S_{n-1})^n < \lim_{n \rightarrow \infty}\left(1+\frac{2}{n}+\frac{4}{n(n-1)}+\frac{6(n-5)}{n(n-1)(n-2)}\right)^n.$$

$$\Rightarrow\exp\left(\lim_{n \rightarrow \infty}n\left(1+\frac{2}{n}-1\right)\right) < L <\exp \left(n \lim_{n \rightarrow \infty} \left(1+\frac{2}{n}+\frac{4}{n(n-1)}+\frac{6(n-5)}{n(n-1)(n-2)}-1 \right)\right).$$ $\Rightarrow e^2 < L< e^2,$ So by sandwich theorem $L=e^2$.

Case (iii) When $N=n-2$: $$S_{n-2}=1+\frac{1}{n\choose 1}+\frac{2}{n\choose 2}+\frac{1}{n\choose 3}+\frac{1}{n\choose 4}+\frac{1}{n\choose 5}+...+\frac{1}{n\choose n-3}.$$ $$\Rightarrow \lim_{n\rightarrow \infty} \left(1+\frac{1}{n}\right)^n <\lim_{n \rightarrow \infty} (S_{n-2})^n < \lim_{n \rightarrow \infty}\left(1+\frac{1}{n}+\frac{4}{n(n-1)}+\frac{6(n-5)}{n(n-1)(n-2)}\right)^n.$$ $$\Rightarrow\exp\left(\lim_{n \rightarrow \infty}n\left(1+\frac{1}{n}-1\right)\right) < L <\exp \left(n \lim_{n \rightarrow \infty} \left(1+\frac{1}{n}+\frac{4}{n(n-1)}+\frac{6(n-5)}{n(n-1)(n-2)}-1 \right)\right).$$ $\Rightarrow e < L< e,$ so $L$ in this case equals $e.$ As pointed by Dr. Hintze, for $N=n-m. m\ge 2$, it can be seen here that $L=e.$