Using the Weierstrass test, show that the series sum
Using the Weierstrass test, show that the series $\displaystyle\sum_{n=2}^{\infty}\ln\!\left(1+\frac{x^2}{n\ln^2n}\right)$ converges uniformly on $[-a,a]$ for any fixed $a>0$.
Explanation of Method
Using $\ln(1+t)\le t$ for $t\ge0$, each term is bounded by $\dfrac{x^2}{n\ln^2n}\le\dfrac{a^2}{n\ln^2n}$ on $[-a,a]$. The majorant series $\sum\dfrac{a^2}{n\ln^2n}$ converges by the Cauchy integral test, since $\displaystyle\int_2^\infty\frac{du}{u\ln^2u}$ converges (substitute $v=\ln u$). The M-test then finishes the job.
- Bound $\ln(1+t)\le t$ to get a term bound $\dfrac{a^2}{n\ln^2n}$ on $[-a,a]$.
- Test convergence of $\sum\dfrac1{n\ln^2n}$ via the integral test.
- Apply the Weierstrass M-test to conclude uniform convergence.
Step 1 · Uniform bound
For $t\ge0$, $\ln(1+t)\le t$ (concavity of $\ln$, tangent line at $t=0$). With $t=\dfrac{x^2}{n\ln^2n}\ge0$: $$ \ln\!\left(1+\frac{x^2}{n\ln^2n}\right) \le \frac{x^2}{n\ln^2n}. $$ On $[-a,a]$, $x^2\le a^2$, so each term is bounded by $M_n=\dfrac{a^2}{n\ln^2n}$, independent of $x$.
Step 2 · Convergence of the majorant
Consider $f(u)=\dfrac1{u\ln^2u}$ for $u\ge2$: positive, continuous, decreasing. By the Cauchy integral test, $\sum_{n=2}^{\infty}f(n)$ converges iff $\displaystyle\int_2^{\infty}\frac{du}{u\ln^2u}$ converges. Substituting $v=\ln u$, $dv=du/u$: $$ \int_2^{\infty}\frac{du}{u\ln^2u} = \int_{\ln2}^{\infty}\frac{dv}{v^2} = \left[-\frac1v\right]_{\ln2}^{\infty} = \frac1{\ln2} < \infty. $$ So $\sum_{n=2}^{\infty}1/(n\ln^2n)$ converges, and hence $\sum M_n=a^2\sum1/(n\ln^2n)<\infty$.
Step 3 · Weierstrass M-test
With $M_n=\dfrac{a^2}{n\ln^2n}$, both M-test conditions hold, so the series converges uniformly and absolutely on $[-a,a]$ for every fixed $a>0$. Since $a$ is arbitrary, this gives uniform convergence on every bounded interval, hence (local uniform convergence) a continuous sum on all of $\mathbb R$.
Summary Table
| Quantity | Value |
|---|---|
| Majorant | $M_n=a^2/(n\ln^2n)$ |
| $\sum1/(n\ln^2n)$ | converges (integral test, $=1/\ln2$ for the integral bound) |
Additional Notes
- This is the most delicate convergence test in the batch: $1/(n\ln^2n)$ decays only barely faster than the divergent $\sum1/(n\ln n)$ (the borderline case), so this problem sits right at the edge of what elementary comparison tests can handle, requiring the integral test's precision.
- The restriction to $[-a,a]$ (rather than all of $\mathbb R$) is essential here, unlike Problems 192, 194, 197, and 198: since the bound uses $x^2\le a^2$, an unbounded $x$-domain would make $\sum M_n$ diverge along with $x^2$ itself, so this result genuinely needs the bounded-interval hypothesis, echoing Problem 183's restriction to bounded intervals for $(1+x/n)^n$.
- This closes out the batch's tour of the Weierstrass M-test (Problems 192–200): together they illustrate every flavor of majorant construction encountered — trigonometric bounds ($|\sin|,|\cos|\le1$), factorial decay, geometric decay, dropped non-negative terms, bounded exponential factors, domain-restricted power bounds, and now a logarithmically refined integral-test bound.
Result
Converges uniformly and absolutely on $[-a,a]$ for every fixed $a>0$: