Infinite Series\Decimal Expansion Proof

Question

Let $(d_k)_{k=1}^\infty$ be a sequence of digits. Then $\sum_{j=1}^{\infty}\limits d_j 10^{-j}$ converges

Could anybody give me a hint on how to begin proving this?

Answer 1

Assumption: $(d_k) \in \{0,1,2,3,4,5,6,7,8,9\}$.

Consider the sequence of partial sums $(\sum_{j=1}^k d_j \cdot 10^{-j}: k \in \mathbb{N})$. Clearly the partial sums are bounded $\forall k \in \mathbb{N}$ as shown below:

$ \hspace{2mm} \sum_{j=1}^k 0 \cdot 10^{-j}\leq\sum_{j=1}^k d_j \cdot 10^{-j} \leq \sum_{j=1}^k 9 \cdot 10^{-j} \hspace{1mm} $

Furthermore, the sequence is monotone increasing as shown below:

$\sum_{j=1}^{k+1} d_j \cdot 10^{-j} - \sum_{j=1}^k d_j \cdot 10^{-j} = d_{k+1}\cdot 10^{k+1} \geq 0$

Then by the Monotone Convergence Theorem, the sequence converges.