Examine the convergence of the series: sum from n=1
Examine the convergence of the series: $$ \sum_{n=1}^{\infty}\left(\frac{1+\cos n}{2+\cos n}\right)^{2n-\ln n}. $$
Explanation of Method
The root test reduces the exponent to $2-\frac{\ln n}n\to2$, so the whole question is whether the base $\frac{1+\cos n}{2+\cos n}$, raised to the power $2$, stays below $1$. Rewriting the base as $1-\frac1{2+\cos n}$ shows it is always bounded above by $\frac23$, since $\cos n\in[-1,1]$ forces $2+\cos n\in[1,3]$ — giving $\rho\le(2/3)^2=4/9<1$.
- Apply the root test and simplify the exponent to $2-\frac{\ln n}n\to2$.
- Bound the base $\frac{1+\cos n}{2+\cos n}$ uniformly by $\frac23$.
- Combine to get $\rho\le4/9<1$ and conclude convergence.
Step 1 · Apply the root test
With $a_n=\left(\dfrac{1+\cos n}{2+\cos n}\right)^{2n-\ln n}$:
$$ \sqrt[n]{a_n} = \left(\frac{1+\cos n}{2+\cos n}\right)^{2-\frac{\ln n}n} \longrightarrow \left(\frac{1+\cos n}{2+\cos n}\right)^{2}, $$
since $\frac{\ln n}n\to0$.
Step 2 · Bound the base
Rewrite $\dfrac{1+\cos n}{2+\cos n} = 1-\dfrac1{2+\cos n}$. Since $-1\le\cos n\le1$, we have $1\le2+\cos n\le3$, so $\dfrac13\le\dfrac1{2+\cos n}\le1$, giving:
$$ 0 \le \frac{1+\cos n}{2+\cos n} \le \frac23 < 1 \qquad \text{for every } n. $$
Step 3 · Combine and conclude
$$ \limsup_{n\to\infty}\sqrt[n]{a_n} \le \left(\frac23\right)^2 = \frac49 < 1. $$
By Cauchy's root test, the series converges. (The bound $\frac23$ is achieved when $\cos n$ is arbitrarily close to $1$, which occurs infinitely often by density of $\{n\bmod2\pi\}$, so the estimate $4/9$ is sharp as a $\limsup$.)
Summary Table
| Quantity | Value |
|---|---|
| Exponent limit | $2$ |
| Base bound | $\dfrac{1+\cos n}{2+\cos n}\le\dfrac23$ |
| $\limsup\sqrt[n]{a_n}$ | $4/9$ |
| Conclusion | Converges |
Additional Notes
- The subtraction of $\ln n$ from $2n$ in the exponent is a red herring for convergence purposes: since $\ln n$ grows so much slower than $n$, the exponent's per-term contribution $2-\frac{\ln n}n$ still converges to the clean integer value $2$, exactly as if the $\ln n$ term weren't there at all.
- This problem shares its Weyl-equidistribution flavor with Problems 69 and 82: the oscillating factor $\cos n$ never actually reaches its extreme value $1$ exactly at integer arguments, but comes arbitrarily close infinitely often, which is enough to make the bound $2/3$ tight without needing exact equality.
- Because the base is always strictly less than $1$ (in fact $\le2/3$) and the exponent tends to a fixed positive value, this series converges "geometrically" in spirit, similar to Problem 74's $\left(\frac{n-2}{n+2}\right)^{n(n+1)}\to e^{-4}$, though here the limiting ratio is algebraic ($4/9$) rather than transcendental.
Result
The series converges: