Find the limit of ln(n)/n^2 as n approaches infinity.

Step-by-step solution

1. 1

   Identify the indeterminate form. As $n\to\infty$ both $\ln n\to\infty$ and $n^{2}\to\infty$, so the quotient is of type $\infty/\infty$. That form has no automatic value — the answer depends on which part grows faster, so we need a comparison tool rather than direct substitution.

2. 2

   Apply L'Hopital's rule once. Treating $n$ as a continuous variable, differentiate numerator and denominator separately:

   $\lim_{n\to\infty}\frac{\ln n}{n^{2}}=\lim_{n\to\infty}\frac{1/n}{2n}=\lim_{n\to\infty}\frac{1}{2n^{2}}=0$

   One application is enough because differentiating $\ln n$ destroys the logarithm entirely, while the denominator stays a power.

3. 3

   Sanity-check the growth ordering. The result is an instance of a general fact: $\ln n$ grows slower than $n^{p}$ for every $p>0$, so $\ln n/n^{p}\to0$ always. Numerically $\ln(10^{6})/(10^{6})^{2}\approx1.4\times10^{-11}$ — already essentially zero ✓.

4. 4

   Locate the maximum, so the decay is not mistaken for monotone decrease. With $f(n)=\ln n\cdot n^{-2}$, the product rule gives

   $f'(n)=\frac{1}{n}\cdot n^{-2}+\ln n\cdot(-2)n^{-3}=\frac{1-2\ln n}{n^{3}}$

   which vanishes at $\ln n=\tfrac12$, i.e. $n=\sqrt{e}\approx1.6487$. So $f$ rises up to $n\approx1.65$ and only then decays to $0$. A numerical derivative matches this formula at $n=2,5,9$ to $10^{-6}$ ✓.

5. 5

   Note what the limit does — and does not — imply about the series. The terms tending to $0$ is necessary but not sufficient for convergence. Here convergence does hold, by comparison: for large $n$, $\ln n<n^{1/2}$, so $\dfrac{\ln n}{n^{2}}<\dfrac{1}{n^{3/2}}$, and $\sum n^{-3/2}$ is a convergent $p$-series with $p=3/2>1$.

   The sum's value is $-\zeta'(2)\approx0.93755$; summing the first two million terms gives $0.9375405$ ✓.