Consider the sequence of rational numbers ${{q_k}}_{k\geq1}$ where $$q_k=\sum_{n=1}^k 1/10^{n^2}$$ i.e., the sequence is $q_1=0.1$, $q_2=0.1001$, $q_3=0.100100001$ etc. So is this sequence bounded, Cauchy and convergent in $Q$? I proved that this sequence is bounded (as it is both bounded below and above). Also, I know that this sequence is not convergent in $Q$ and it converges to $0.100100001...$ but I could not determine how and I also was not able to find the method to prove this sequence Cauchy. p.s. I know this same question has been asked before but the answer over there does not help me completely. Please help.
2026-04-05 19:37:36.1775417856
How to prove that the following sequence is Cauchy?
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Cauchy property is quite easy: $q_{k+j}-q_j =\sum\limits_{k=j+1}^{j+k}\frac 1{10^{k^{2}}}<\sum\limits_{k=j+1}^{j+k}\frac 1{10^{k}} $. Now apply the formula for a geometric sum (or just use convergence of the series $\sum \frac 1 {10^{k}}$.