Find the limit of the sequence $\{.1, .101, .101001, \ldots,\}$.

Reference:- Foundations of Functional Analysis, S. Ponnusammy, $\alpha-science-2002$

Consider the sequence $\{.1, .101, .101001, \ldots\}$, prove that given sequence is Cauchy and converges to a irrational number. find the limit of the sequence.

I tried to find the generalised formula for the sequence pattern. I failed to do. $\left\{\frac{1}{10}, \frac{1}{10}+\frac{1}{10^3}, \frac{1}{10}+\frac{1}{10^3}+\frac{1}{10^6}, \ldots\right\}$. It doesn't form any geometric series. How do I prove the sequence is Cauchy without getting the rigourous expression for the $n$-th term of the given sequence? how to find the limit of the sequence?

• These three initial elements of the sequence do not give any information on how it continues. Are you sure this is the full problem statement? Commented Oct 30, 2017 at 15:14
• The $1$'s seem to be (I would guess) at the $1,3,6,10,\ldots$th position (so the $\frac{n(n+1)}{2}$th position for $n=1,2,3,\ldots$). If so then we can write $x = \sum_{n=1}^\infty 10^{-\frac{n(n+1)}{2}}$. Commented Oct 30, 2017 at 15:15
• yes. I will post the picture of the problem.
– user464147
Commented Oct 30, 2017 at 15:15
• You can write a formula (for what the sequence is, I assume, supposed to be): $a_n=\sum_{i=1}^n 10^{\sum_{k=1}^i k}$, but that's hardly helpful. Just take the difference between two elements and look at its decimal expansion. The estimates to make are trivial, you don't need any deep understanding of the sequence beyond the fact that its decimal expansion stabilises. Commented Oct 30, 2017 at 15:17
• Not completely on topic, but this video might be interesting. Commented Oct 30, 2017 at 15:19

You can prove it is Cauchy directly from the definition. If I give you an $\epsilon \gt 0$ you just need to find an $N$ such that no two terms beyond $N$ differ by more than $\epsilon$. All the changes after $k$ add up to less than $10^{-k}$
Hint: If $a$ and $b$ are positive, smaller than $1$, and they agree up to $k$ decimal places, then $\lvert a-b\rvert<\ldots$
• A similar sequence could be with folowing general term: $\frac{101 . (10^n +1) - 100}{10^{2n}}$ which hods true for 3rd and further terms. The limit of this is: $Lim_{n→∞ } \frac{101 . (10^n +1) - 100}{10^{2n}}=Lim_{n→∞ }\frac{101 . (1/10^n +1) - 100/10^n}{10^{n}}=0$ Commented Oct 30, 2017 at 20:51