Knowing that sum from n=1 to infinity 1/(n^()alpha)
Knowing that $\sum_{n=1}^{\infty}\frac{1}{n^\alpha}$ converges for $\alpha>1$ and diverges for $\alpha\le1$, determine the convergence of: $$ \sum_{n=3}^{\infty} \ln\frac{\cosh\frac{\pi}{n}}{\cos\frac{\pi}{n}}. $$
Explanation of Method
Both $\cosh x$ and $\cos x$ share the Taylor expansion $1\pm\frac{x^2}2+\frac{x^4}{24}+\cdots$, differing only in the sign of the $x^2$ term. After taking logarithms and subtracting, the leading $O(x^4)$ pieces cancel and the surviving leading term is exactly $x^2$ — here $\left(\frac\pi n\right)^2$ — placing the series comfortably in the convergent regime.
- Taylor expand $\cosh(\pi/n)$ and $\cos(\pi/n)$ to fourth order.
- Take logarithms of each and subtract, tracking the leading-order cancellation.
- Apply the limit comparison test against $\sum 1/n^2$.
Step 1 · Taylor expand
With $x=\frac\pi n$: $\cosh x = 1+\frac{x^2}2+\frac{x^4}{24}+O(x^6)$, $\cos x = 1-\frac{x^2}2+\frac{x^4}{24}+O(x^6)$.
Step 2 · Expand the logarithms and subtract
Using $\ln(1+y)=y-\frac{y^2}2+O(y^3)$ on each factor and simplifying (the $x^4$ contributions combine with the $-y^2/2$ correction), both logarithms work out to:
$$ \ln(\cosh x) = \frac{x^2}2-\frac{x^4}{12}+O(x^6), \qquad \ln(\cos x) = -\frac{x^2}2-\frac{x^4}{12}+O(x^6). $$
Subtracting, the $x^4$ terms cancel exactly:
$$ a_n = \ln(\cosh x)-\ln(\cos x) = x^2+O(x^6) = \frac{\pi^2}{n^2}+O\!\left(\frac1{n^6}\right). $$
Step 3 · Limit comparison test
With $b_n=\dfrac1{n^2}$:
$$ \lim_{n\to\infty}\frac{a_n}{b_n} = \lim_{n\to\infty} n^2\ln\frac{\cosh(\pi/n)}{\cos(\pi/n)} = \pi^2. $$
Positive and finite; since $\sum\frac1{n^2}$ converges, by the limit comparison test the original series converges too.
Summary Table
| Quantity | Value |
|---|---|
| General term $a_n$ | $\sim\dfrac{\pi^2}{n^2}$ |
| Limit comparison ratio (vs. $1/n^2$) | $\pi^2$ |
| Conclusion | Converges |
Additional Notes
- The essential structural fact is that $\cosh x$ and $\cos x$ are "mirror" functions — replacing $x$ with $ix$ turns one into the other — so their Taylor coefficients agree in magnitude but alternate in sign on the $x^2$ term while matching on the $x^4$ term, producing the clean cancellation seen in Step 2.
- Since $\cosh x>1>\cos x$ for $0<x<\pi/2$ (true here for $n\ge3$, giving $x=\pi/n\le\pi/3$), the ratio $\cosh x/\cos x>1$, so every term of the series is positive and $\ln$ of it is well-defined and positive.
- This problem parallels Problem 35's cancellation of the exponential's linear term via $a^{1/n}+a^{-1/n}$: in both cases, combining a function with its "even-symmetrized" counterpart cancels the leading term and boosts convergence to the next order.
Result
The series converges: