Using the Cauchy criterion, prove that the following
Using the Cauchy criterion, prove that the following series converges: $$ a_0 + \sum_{i=1}^{\infty} \frac{a_i}{b^i} \qquad (|a_i| < b,\; b>1). $$
Explanation of Method
The bound $|a_i|<b$ turns each term $|a_i/b^i|$ into something smaller than $1/b^{i-1}$, which is a term of a convergent geometric series (since $b>1$). Bounding the Cauchy tail sum of the original series by the tail of that geometric series, and using that geometric tails shrink to zero, produces exactly the estimate the Cauchy criterion requires.
- Bound each term using $|a_i|<b$.
- Sum the resulting geometric bound over a Cauchy tail interval $[n+1,m]$.
- Show the bound can be made arbitrarily small, verifying the Cauchy criterion.
Step 1 · Set up the Cauchy tail sum
Write $c_0=a_0$, $c_i=a_i/b^i$ for $i\ge1$. For $m>n\ge1$, by the triangle inequality:
$$ \left|\sum_{i=n+1}^{m} \frac{a_i}{b^i}\right| \le \sum_{i=n+1}^{m} \frac{|a_i|}{b^i}. $$
Step 2 · Bound using $|a_i|<b$
Since $|a_i|<b$, each term satisfies $\dfrac{|a_i|}{b^i} < \dfrac{b}{b^i}=\dfrac{1}{b^{i-1}}$. Reindexing with $j=i-1$:
$$ \sum_{i=n+1}^{m}\frac{1}{b^{i-1}} = \sum_{j=n}^{m-1}\frac{1}{b^j} = \frac{1}{b^n}\cdot\frac{1-b^{-(m-n)}}{1-\frac1b} < \frac{1}{b^n}\cdot\frac{b}{b-1} = \frac{1}{b^{n-1}(b-1)}. $$
Step 3 · Apply the Cauchy criterion
So $\left|\sum_{i=n+1}^{m}\frac{a_i}{b^i}\right| < \dfrac{1}{b^{n-1}(b-1)}$. Given $\varepsilon>0$, since $b>1$ implies $b^{n-1}\to\infty$, choose $N$ with $\dfrac1{b^{N-1}(b-1)}<\varepsilon$. Then for all $m>n\ge N$:
$$ \left|\sum_{i=n+1}^{m}\frac{a_i}{b^i}\right| < \frac{1}{b^{n-1}(b-1)} \le \frac{1}{b^{N-1}(b-1)} < \varepsilon. $$
This is exactly the Cauchy criterion, so the series converges.
Summary Table
| Quantity | Bound |
|---|---|
| $|a_i/b^i|$ | $< 1/b^{i-1}$ |
| Tail sum $\left|\sum_{i=n+1}^m a_i/b^i\right|$ | $< \dfrac{1}{b^{n-1}(b-1)}$ |
| Conclusion | Converges (Cauchy criterion, dominated by geometric series) |
Additional Notes
- This is essentially a comparison test disguised as a Cauchy-criterion proof: the series is dominated term-by-term by the convergent geometric series $\sum 1/b^{i-1}$, and convergence of that dominating series directly forces convergence of the tail sums here.
- The specific structure $|a_i|<b$ with base-$b$ denominators is exactly the setting of base-$b$ digit expansions (e.g. decimal expansions when $b=10$), where each "digit" $a_i$ is bounded by the base, guaranteeing the expansion converges to a well-defined real number.
- The bound is uniform in the sense that it depends only on $n$ and $b$, not on the specific values of the $a_i$ (only on the bound $|a_i|<b$), so the same estimate works for every admissible choice of coefficients $\{a_i\}$.
Result
The series converges by the Cauchy criterion, since: