Infinite sum of rationals is irrational 
a) Let $x$ be a number between $0$ and $1$.  Let $a_1$ be the smallest positive integer such that $x_1=x-a_1^{-1}\geq 0$, let $a_2$ be the smallest positive integer such that $x_2=x_1-a_2^{-1}\geq 0$, etc.  Show that this leads to a finite expansion $$x=\frac{1}{a_1}+\frac{1}{a_2}+\dots+\frac{1}{a_n}$$ (that is, that $x_{n+1}=0$ for some $n$) if and only if $x$ is rational.
b) Show that if the integers $1<b_1<b_2<\cdots$ increase so rapidly that $$\frac{1}{b_{k+1}}+\frac{1}{b_{k+2}}+\dots<\frac{1}{b_k-1}-\frac{1}{b_k}~~\text{for }k\geq 1,$$ then the number $\sum b_k^{-1}$ is irrational.  Use this then to prove that $\sum\limits_{k=0}^\infty (2^{3^k}+1)^{-1}$ is irrational.

I solved (a), but (b) is a little difficult.
This is my way to (b)
suppose $\sum b_k^{-1}$ is rational and I want this is a contradiction.
using the result of (a), 1/b_k−1 can be expressed with finite rational expansion.
1/b_k-1>$\sum b_k^{-1}$ which is rational
so I put 1/b_k-1=$\sum b_k^{-1}$+a/b which b is not zero.
Now, What should I do?
how to solve (b)?
 A: To solve (b), we can use (a). Let $x=\sum b_k^{-1}$, and for $n\ge 1$, let $x_n=\sum_{k>n} b_k^{-1}$.
We have to show $b_{n + 1}$ is the smallest integer s.t. $x_n - b_{n+1}^{-1}\ge 0$ for all $n$. 
Let $n\in\mathbb{N}$. On the one hand, $x_n - b_{n+1}^{-1}=x_{n+1}$ is indeed positive. On the other hand
\begin{eqnarray}
x_{n} - \frac{1}{b_{n+1} - 1} & = & \left(\frac{1}{b_{n+1}} - \frac{1}{b_{n+1} - 1}\right) + \sum_{k>n+1}\frac{1}{b_k}\\
& < & \left(\frac{1}{b_{n+1}} - \frac{1}{b_{n+1} - 1}\right) + \left(\frac{1}{b_{n+1} - 1} - \frac{1}{b_{n+1}}\right)\\
& = & 0
\end{eqnarray}
thus, $b_{n + 1}$ is the smallest integer s.t. $x_n - b_{n+1}^{-1}\ge 0$. Hence, $x$ is of the form shown in (a), with all the terms $x_n >0$, thus it is irrational.
To prove $x=\sum_k (2^{3^k}+ 1)^{-1}$ is irrational, note that for all $k$, 
$$
\frac{1}{2^{3^k}} - \frac{1}{2^{3^k} + 1} = \frac{1}{2^{3^k}}\left
(1 - \frac{2^{3^k}}{2^{3^k} + 1}\right) = \frac{1}{2^{3^k}}\left
(\frac{1}
{2^{3^k} + 1}\right) := m_k
$$
while, on the other hand,
$$
\sum_{n>k}\frac{1}{2^{3^n} + 1} < \frac{1}{2^{3^{k+1}} + 1} + \sum_{n\ge 3^{k+2}}\frac{1}{2^n} = \frac{1}{2^{3^{k+1}} + 1} + \frac{2}{2^{3^{k+2}}} := n_k
$$
We can show that for all $k, m_k>n_k$. Hence by part (b) we conclude $x$ is irrational.
