convergence of a sequence (partial fractions) Let's define {${x_n}$} as $x_1=0.1, x_2=0.101, x_3=0.101001,....$ . Then we need to find out if the sequence convergences or not and the limit. 
This is how i proceeded. $x_1=\frac{1}{10}, x_2=\frac{1}{10}+\frac{1}{1000}, x_3=\frac{1}{10}+\frac{1}{1000}+\frac{1}{1000000},..$ and so on. This is an increasing sequence and bounded above. Hence it converges. But i am unable to find the limit
 A: It looks like
$x = 10^{-1}+10^{-3}
+10^{-6}+10^{-10} + ...
=\sum_{n=1}^{\infty} 10^{-t_n}
$,
where
$t_n = \dfrac{n(n+1)}{2}$
is the $n$-th triangular number.
This is a value of the
$q$-series Theta function
$\psi(q)
=\sum_{n=0}^{\infty} q^{n(n+1)/2}
$
which I found discussed on page 6 of
"Number Theory in the Spirit of Ramanujan"
by Bruce Berndt.
In later chapters,
it is shown how
this function is related to
hypergeometric functions,
elliptic functions,
and Eisenstein series.
This is also related to
the Jacobi theta series
$θ_2(q) = 2q^{1/4}
\sum_{n≥0}
q^{n(n+1)}
$.
It turns out that $x$ is transcendental.
After some searching,
I found this reference:
"Introduction to Algebraic Independence Theory, Issue 1752"
 edited by Patrice Philippon,
published by Springer.
On page 30,
it states (slightly edited) that
Corollary 1.9.
Let $f(z)$ be one of the functions
$\theta_2(z)$,
$\theta_3(z)$,
$\theta_4(z)$.
For any algebraic number $q$
with $0 < |q| < 1$,
the values
$f(q), f'(q), f''(q)$
are algebraically independent.
In particular,
$f(q)$ is transcendental.
If we let
$q = 1/\sqrt{10}$,
$\theta_2(q)
=\dfrac{2}{10^{1/8}}
\sum_{n≥0}
10^{-n(n+1)/2}
=\dfrac{2}{10^{1/8}} x
$
is transcendental,
so $x$ is also.
A: A trivial way to write the limit is simply as $0.101001000100001\ldots$. The limit is irrational because the sequence of digits never repeats.
A: The sequence of rational numbers $0.1, 0.101, 0.101001, 0.1010010001, \ldots$ converges to $0.101001000100001000\ldots$ However, by the Completeness Axiom (the least upper bound property or $\mathbb R$) every nonempty $S \subseteq \mathbb R$ with an upper bound must have a least upper bound (supremum) in $\mathbb R$. Moreover, $\sup S$ is unique so every nonempty $S \subseteq \mathbb R$ with an upper bound must have the least upper bound in $\mathbb R$. Therefore $0.101001000100001000\ldots = \sup\{0.1, 0.101, 0.101001, 0.1010010001, \ldots\}$. Overall, the limit is $0.101001000100001000\ldots$ but you can write it as $\sup\{0.1, 0.101, 0.101001, 0.1010010001, \ldots\}$ if you like.
