MHB Calculate Upper Bound for $\displaystyle a_{n}$ in Binomial Limit Evaluation

[juantheron]

Evaluation of $\displaystyle \lim_{n\rightarrow \infty}\sum^{n}_{k=0}\frac{1}{\binom{n}{k}}$ is

I have tried like this way::

Let $\displaystyle a_{n} = \sum^{n}_{k=0}\frac{1}{\binom{n}{k}} = \frac{1}{\binom{n}{0}}+\frac{1}{\binom{n}{1}}+\sum^{n-2}_{k=2}\frac{1}{\binom{n}{k}}+\frac{1}{\binom{n}{n-1}}+\frac{1}{\binom{n}{n}}\geq 2+\frac{2}{n}$

Now I did not understand How can I calculate upper bound for $a_{n}$

Thanks

 

[Opalg]

Evaluation of $\displaystyle \lim_{n\rightarrow \infty}\sum^{n}_{k=0}\frac{1}{\binom{n}{k}}$ is

I have tried like this way::

Let $\displaystyle a_{n} = \sum^{n}_{k=0}\frac{1}{\binom{n}{k}} = \frac{1}{\binom{n}{0}}+\frac{1}{\binom{n}{1}}+\sum^{n-2}_{k=2}\frac{1}{\binom{n}{k}}+\frac{1}{\binom{n}{n-1}}+\frac{1}{\binom{n}{n}}\geq 2+\frac{2}{n}$

Now I did not understand How can I calculate upper bound for $a_{n}$

Notice that ${n\choose k} \geqslant {n\choose 2} = \frac12n(n-1)$ for $2\leqslant k\leqslant n-2$. So the sum $$\sum^{n-2}_{k=2}\frac{1}{n\choose k}$$ has $n-3$ terms, each of which is $\leqslant \frac2{n(n-1)}.$ That gives you an estimate for the upper bound of $a_n.$