Using the Cauchy criterion, test the uniform convergence
Using the Cauchy criterion, test the uniform convergence of the series $\displaystyle\sum_{n=2}^{\infty}\dfrac1{(x+n-1)(x+n)}$ on $[0,a]$.
Explanation of Method
Partial fractions turn the series into a telescoping sum, so the tail beyond term $m$ collapses exactly to $1/(x+m)$, with no approximation needed. Since $x\ge0$ on $[0,a]$, this exact remainder is bounded above by $1/m$ regardless of $x$, giving a clean, sharp Cauchy-criterion bound.
- Decompose the term via partial fractions and identify the telescoping structure.
- Compute the exact tail sum (remainder) in closed form.
- Bound the remainder uniformly on $[0,a]$ and confirm the Cauchy criterion.
Step 1 · Partial fractions
$\dfrac1{(x+n-1)(x+n)}=\dfrac1{x+n-1}-\dfrac1{x+n}$.
Step 2 · Exact remainder
$$ \sum_{k=m+1}^n\frac1{(x+k-1)(x+k)} = \sum_{k=m+1}^n\left(\frac1{x+k-1}-\frac1{x+k}\right) = \frac1{x+m}-\frac1{x+n}. $$ As $n\to\infty$, the remainder is exactly $R_m(x)=\dfrac1{x+m}$.
Step 3 · Uniform bound and conclusion
For $x\in[0,a]$: $\dfrac1{x+m}$ is decreasing in $x$, so it is maximized at $x=0$: $$ \sup_{x\in[0,a]}\left|\sum_{k=m+1}^n\frac1{(x+k-1)(x+k)}\right| \le \frac1{0+m} = \frac1m \to 0. $$ Given $\epsilon>0$, choose $N>1/\epsilon$; the Cauchy criterion then holds uniformly for all $n>m\ge N$ and all $x\in[0,a]$.
Summary Table
| Quantity | Value |
|---|---|
| Exact remainder $R_m(x)$ | $1/(x+m)$ |
| $\sup_{[0,a]}R_m(x)$ | $1/m\to0$ |
| Sum $S(x)$ | $1/(x+1)$ |
Additional Notes
- This is the functional (parametrized) analogue of Problem 149's numeric telescoping identity $\sum_{m,n\ge2}1/m^n=1$: the same partial-fraction telescoping mechanism appears, here producing an exact remainder rather than requiring a separate estimate.
- Because the remainder $1/(x+m)$ is computed exactly (not just bounded), this problem offers one of the cleanest possible Cauchy-criterion verifications in this batch — no approximation, comparison series, or asymptotic estimate is needed at any step.
- The Weierstrass M-test gives the identical conclusion via $M_n=1/[(n-1)n]$ (bounding at $x=0$), with $\sum M_n=1$ — a useful cross-check confirming the sharper, exact Cauchy-criterion computation above.
Result
The series converges uniformly on $[0,a]$: