Using the Cauchy criterion, prove that the following
Using the Cauchy criterion, prove that the following series converges: $$ \sum_{n=1}^{\infty} \frac{\cos(nx) - \cos((n+1)x)}{n}. $$
Explanation of Method
Reindexing the shifted cosine term turns the tail sum into three pieces: two boundary terms and a middle sum whose coefficient $\frac1k-\frac1{k-1}$ is itself a telescoping expression. Bounding $|\cos|\le1$ throughout and summing the telescoping coefficient exactly (rather than estimating it) produces a clean bound of $\frac{2}{n+1}$ on the whole tail sum — small enough, uniformly in $x$, to satisfy the Cauchy criterion.
- Reindex the shifted cosine term to combine both sums over a common index.
- Bound the result using $|\cos|\le1$ and evaluate the telescoping coefficient sum exactly.
- Apply the Cauchy criterion using the resulting bound $\dfrac{2}{n+1}$.
Step 1 · Reindex and combine
Splitting and reindexing $j=k+1$ in the second sum of $\sum_{k=n+1}^{m}\frac{\cos(kx)-\cos((k+1)x)}{k}$, then regrouping, gives:
$$ \sum_{k=n+1}^{m}\frac{\cos(kx)-\cos((k+1)x)}{k} = \frac{\cos((n+1)x)}{n+1} - \sum_{k=n+2}^{m}\cos(kx)\left(\frac1{k-1}-\frac1k\right) - \frac{\cos((m+1)x)}{m}. $$
Step 2 · Bound using $|\cos|\le1$ and telescope the coefficients
Applying the triangle inequality and $|\cos(\cdot)|\le1$:
$$ \left|\sum_{k=n+1}^{m}\frac{\cos(kx)-\cos((k+1)x)}{k}\right| \le \frac{1}{n+1} + \sum_{k=n+2}^{m}\frac{1}{k(k-1)} + \frac1m. $$
Using $\frac1{k(k-1)}=\frac1{k-1}-\frac1k$, the middle sum telescopes exactly: $\sum_{k=n+2}^{m}\left(\frac1{k-1}-\frac1k\right) = \frac1{n+1}-\frac1m$. Substituting:
$$ \left|\sum_{k=n+1}^{m}\frac{\cos(kx)-\cos((k+1)x)}{k}\right| \le \frac{1}{n+1}+\frac1{n+1}-\frac1m+\frac1m = \frac{2}{n+1}. $$
Step 3 · Apply the Cauchy criterion
Given $\varepsilon>0$, choose $N>\frac2\varepsilon-1$, so $\frac2{N+1}<\varepsilon$. Then for all $m>n\ge N$:
$$ \left|\sum_{k=n+1}^{m}\frac{\cos(kx)-\cos((k+1)x)}{k}\right| \le \frac{2}{n+1} \le \frac{2}{N+1} < \varepsilon. $$
This is the Cauchy criterion; it holds with the same $N$ for every $x$, so the series converges — uniformly in $x\in\mathbb R$.
Summary Table
| Quantity | Bound |
|---|---|
| Tail sum $\left|\sum_{k=n+1}^{m} a_k\right|$ | $\le \dfrac{2}{n+1}$ |
| Dependence on $x$ | None (bound is uniform) |
| Conclusion | Converges, uniformly in $x$ |
Additional Notes
- The general term itself, $\frac{\cos(nx)-\cos((n+1)x)}{n}$, is not obviously telescoping in $n$ the way earlier problems' terms were; it was the reindexing step that revealed a hidden telescoping structure in the coefficients $\frac1{k(k-1)}$, not in the cosines themselves.
- Because the bound $\frac2{n+1}$ has no dependence on $x$, the same argument that proves pointwise convergence also proves uniform convergence — a stronger and often more useful conclusion, analogous to what happened in Problem 24.
- Trigonometric identities like $\cos(nx)-\cos((n+1)x) = 2\sin\left(\frac{(2n+1)x}{2}\right)\sin\left(\frac x2\right)$ offer an alternative route to bounding this series, but the reindexing/telescoping approach used here avoids introducing extra trigonometric machinery.
Result
The series converges for all $x\in\mathbb{R}$, uniformly, by the Cauchy criterion: