Now I am proving the number $$ \sum_{k=1}^{\infty}\frac{9}{10^{\frac{k(k+1)}{2}}}=0.90900900090... $$ is irrational.
Here I use a similar method to the proof of e is irrational by Joseph Fourier. My attempt is as follows.
My Attempt: Assume that it is a rational number. Then there exists $ p,q\in \mathbb{Z} $ (with $ q\neq 0 $) such that $$ \sum_{k=1}^{\infty}\frac{9}{10^{\frac{k(k+1)}{2}}}=\frac{p}{q} .$$
Choose $ n\in \mathbb{N} $ such that $ 10^{n}>q+1 $ and $ n=\frac{k_{0}(k_{0}+1)}{2} $ for some $ k_{0}\in \mathbb{N} $.
Put $$ x=q10^{n}\left( \frac{p}{q}-\sum_{k=1}^{k_{0}}\frac{9}{10^{\frac{k(k+1)}{2}}}\right). $$
Then $$x=10^{n}p-q10^{n}\sum_{k=1}^{k_{0}}\frac{9}{10^{\frac{k(k+1)}{2}}}$$ and hence $x$ is an integer.
Clearly $$x=q10^{n}\sum_{k=k_{0}+1}^{\infty}\frac{9}{10^{\frac{k(k+1)}{2}}}>0.$$
Now I need to show $x<1$. But I can't seem to obtain that $x<1$. My attempt for this part is as follows.
$$x=q10^{n}\sum_{k=k_{0}+1}^{\infty}\frac{9}{10^{\frac{k(k+1)}{2}}}<q10^{n}\sum_{k=k_{0}+1}^{\infty}\frac{9}{10^{n+k}}<q\sum_{k=k_{0}+1}^{\infty}\frac{9}{10^{k}}=\frac{q}{10^{k_{0}}}<...$$
Could anyone please give me some help to show $x<1$ ?(if my choice for $x$ is correct. if it is wrong then, what is the suitable choice for $x$ ?)
Any hints/ideas are much appreciated.
If you choose $n=k_0(k_0+1)/2$ with $10^{k_0}>q$ instad of $10^n>q$ you will get the desorden inequalities.