Using Dirichlet's test, prove the convergence of: sum
Using Dirichlet's test, prove the convergence of: $$ \sum_{n=2}^{\infty} \frac{1}{\ln^2 n} \cos \frac{\pi n^2}{n+1}. $$
Explanation of Method
The argument $\frac{\pi n^2}{n+1}$ simplifies via polynomial division to $\pi n-\frac{\pi n}{n+1}$, and since $\cos(\pi n-\theta)=(-1)^n\cos\theta\cdot(-1)$... more precisely $\cos\left(\pi n-\frac{\pi n}{n+1}\right)=(-1)^n\cos\left(\frac{\pi n}{n+1}\right)$, reducing the series to a simple alternating series (times a slowly-varying cosine factor near $-1$) whose Dirichlet-test convergence follows immediately from the bounded partial sums of $(-1)^n$.
- Simplify $\cos\frac{\pi n^2}{n+1}=(-1)^n\cos\left(\frac{\pi n}{n+1}\right)$.
- Split into a leading alternating term and a correction converging absolutely.
- Apply Dirichlet's test to the leading term and absolute convergence to the correction.
Step 1 · Simplify the cosine argument
$\dfrac{n^2}{n+1}=n-\dfrac n{n+1}$, so $\dfrac{\pi n^2}{n+1}=\pi n-\dfrac{\pi n}{n+1}$. Using $\cos(\pi n-\theta)=\cos(\pi n)\cos\theta+\sin(\pi n)\sin\theta=(-1)^n\cos\theta$ (since $\sin\pi n=0$):
$$ \cos\frac{\pi n^2}{n+1} = (-1)^n\cos\left(\frac{\pi n}{n+1}\right). $$
Step 2 · Split into leading and correction terms
Since $\frac{\pi n}{n+1}=\pi-\frac\pi{n+1}\to\pi$: $\cos\left(\frac{\pi n}{n+1}\right)=-\cos\left(\frac\pi{n+1}\right)=-1+\dfrac{\pi^2}{2(n+1)^2}+O(n^{-4})$. So the series splits as:
$$ \sum_{n=2}^{\infty}\frac{(-1)^n}{\ln^2n}\left(-1+\frac{\pi^2}{2(n+1)^2}+O(n^{-4})\right) = -\sum_{n=2}^{\infty}\frac{(-1)^n}{\ln^2n} + \sum_{n=2}^{\infty}\frac{(-1)^n}{\ln^2n}\cdot\frac{\pi^2}{2(n+1)^2}+\cdots $$
Step 3 · Apply Dirichlet's test and absolute convergence
For the leading term: $a_n=(-1)^n$ has $|A_N|\le1$ (bounded), and $b_n=1/\ln^2n$ is decreasing (since $f'(x)=-\frac2{x(\ln x)^3}<0$) with $b_n\to0$, so Dirichlet's test gives convergence of $\sum(-1)^n/\ln^2n$. For the correction: $\left|\frac{(-1)^n}{\ln^2n}\cdot\frac{\pi^2}{2(n+1)^2}\right|\le\frac{\pi^2}{2(n+1)^2}$, and $\sum1/n^2$ converges, so the correction converges absolutely. The full series converges.
Summary Table
| Component | Convergence mechanism |
|---|---|
| $\sum(-1)^n/\ln^2n$ | Dirichlet's test ($|A_N|\le1$, $b_n\downarrow0$) |
| Correction term | Absolute convergence (comparison with $\sum1/n^2$) |
| Conclusion | Converges |
Additional Notes
- The trigonometric simplification $\cos\left(\pi n-\theta\right)=(-1)^n\cos\theta$ is the same $\pi n$-shift identity that makes many "$\cos(\pi\cdot\text{polynomial in }n)$" series reduce to simple alternating series — always look for a hidden factor of $\pi n$ that can be split off using $\sin\pi n=0$.
- This problem's leading term $\sum(-1)^n/\ln^2n$ is essentially Problem 108's series but with the exponent on $\ln n$ doubled — since $1/\ln^2n$ decreases even more slowly than $1/\ln n$, Dirichlet's test (which doesn't care about the rate) handles it exactly as easily.
- Decomposing a series into a "Dirichlet-test part" plus an "absolutely convergent correction" is a common strategy whenever the oscillating factor is only asymptotically close to a simpler oscillating factor (here $\cos(\pi n/(n+1))\approx-1$) — the correction inherits enough decay to be handled by comparison alone.
Result
The series converges: