Test the absolute and conditional convergence
Test the absolute and conditional convergence of the series: $$ \sum_{n=1}^{\infty} \frac{\cos\frac{n\pi}4}{n^a}. $$
Explanation of Method
The factor $\cos\frac{n\pi}4$ is periodic with period $8$, taking the eight values $\frac{\sqrt2}2,0,-\frac{\sqrt2}2,-1,-\frac{\sqrt2}2,0,\frac{\sqrt2}2,1$ in rotation, never all zero. This bounded, oscillating, non-vanishing pattern is exactly the setting for Dirichlet's test: its partial sums stay bounded via the closed-form cosine-sum identity, so pairing with the monotone-decreasing $n^{-a}$ for $a>0$ gives convergence, while $a>1$ additionally secures absolute convergence via direct comparison.
- Compute the closed-form bound on partial sums of $\cos\frac{n\pi}4$.
- Apply Dirichlet's test for $a>0$.
- Determine the absolute-convergence threshold at $a=1$.
Step 1 · Bound the partial sums
$$ \sum_{n=1}^N\cos\frac{n\pi}4 = \frac{\sin\left(\frac{N\pi}8\right)\cos\left(\frac{(N+1)\pi}8\right)}{\sin(\pi/8)}, \qquad \left|\sum_{n=1}^N\cos\frac{n\pi}4\right|\le\frac1{\sin(\pi/8)}\approx2.61. $$
Step 2 · Apply Dirichlet's test
For $a>0$, $n^{-a}$ decreases to $0$, and Step 1 gives bounded partial sums, so by Dirichlet's test the series converges for all $a>0$.
Step 3 · Absolute convergence threshold
$\left|\cos\frac{n\pi}4\right|$ is not identically zero and, over each period of $8$, sums to a fixed positive constant, so $\sum\left|\cos\frac{n\pi}4\right|/n^a$ compares (up to a constant factor) with $\sum1/n^a$: convergent iff $a>1$. So the series is absolutely convergent for $a>1$, conditionally convergent for $0<a\le1$, and for $a\le0$ the terms don't tend to $0$ (they cycle through nonzero values), so it diverges.
Summary Table
| Range of $a$ | Conclusion |
|---|---|
| $a>1$ | Absolutely convergent |
| $0<a\le1$ | Conditionally convergent |
| $a\le0$ | Diverges |
Additional Notes
- The period-$8$ cosine pattern here is the natural companion to Problem 129's period-$24$ sine series and Problem 116's period-$8$ series: all three follow the identical Dirichlet's-test template, differing only in the specific bounded oscillating factor and its comparison constant.
- Because $\cos\frac{n\pi}4=1$ exactly when $8\mid n$, the "worst case" absolute subsequence recovers a genuine (scaled) $p$-series along that arithmetic progression, which is precisely why the absolute-convergence threshold lands exactly at $a=1$, matching the ordinary $p$-series boundary rather than shifting due to the oscillation.
- This problem is the cosine counterpart of Problem 133's sine series $\sum\sin(nx)/n^a$: both share the identical threshold structure ($a>1$ absolute, $0<a\le1$ conditional), reflecting that any bounded, non-degenerate trigonometric oscillation combined with a decreasing power produces the same qualitative convergence picture.
Result
Absolutely convergent for $a>1$; conditionally convergent for $0<a\le1$: