What would you like to understand?

The distinction lies in the quantifier order. For pointwise convergence, one first fixes a specific point $x$, then chooses $\epsilon>0$, and finds an index $N$ that may depend on both $\epsilon$ and $x$.

Conditions: A sequence of functions $f_n$ and a limit function $f$ are defined on a region.; Pointwise convergence: $\forall x, \forall \epsilon>0, \exists N$ such that $n\ge N \implies |f(x)-f_n(x)|<\epsilon$.; Uniform convergence: $\forall \epsilon>0, \exists N$ such that $n\ge N \implies |f(x)-f_n(x)|<\epsilon$ for all $x$ in the region.

The distinction lies in the quantifier order. For pointwise convergence, one first fixes a specific point $x$, then chooses $\epsilon>0$, and finds an index $N$ that may depend on both $\epsilon$ and $x$.

Conditions: A sequence of functions $f_n$ and a limit function $f$ are defined on a region.; Pointwise convergence: $\forall x, \forall \epsilon>0, \exists N$ such that $n\ge N \implies |f(x)-f_n(x)|<\epsilon$.; Uniform convergence: $\forall \epsilon>0, \exists N$ such that $n\ge N \implies |f(x)-f_n(x)|<\epsilon$ for all $x$ in the region.

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.