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Uniform convergence fails on $(1, \infty)$ because of the singularity of $\zeta(x)$ at $x=1$. The lecture constructs a counterexample by choosing a specific $\epsilon = 0.005$ and letting $x$ depend on $n$ as $x = 1 + \frac{1}{n+1}$.

Conditions: The domain is the open interval $x > 1$.; $f_n(x) = \sum_{k=1}^n \frac{1}{k^x}$ is the sequence of partial sums.; $\zeta(x) = \sum_{k=1}^{\infty} \frac{1}{k^x}$ is the limit function.; The integral test result $\zeta(1 + \frac{1}{n+1}) > n+1$ is assumed.

Uniform convergence fails on $(1, \infty)$ because of the singularity of $\zeta(x)$ at $x=1$. The lecture constructs a counterexample by choosing a specific $\epsilon = 0.005$ and letting $x$ depend on $n$ as $x = 1 + \frac{1}{n+1}$.

Conditions: The domain is the open interval $x > 1$.; $f_n(x) = \sum_{k=1}^n \frac{1}{k^x}$ is the sequence of partial sums.; $\zeta(x) = \sum_{k=1}^{\infty} \frac{1}{k^x}$ is the limit function.; The integral test result $\zeta(1 + \frac{1}{n+1}) > n+1$ is assumed.