Why does the sequence of partial sums $f_n(x)$ fail to converge uniformly to zeta(x) on the interval (1, infinity)?

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.

Reasoning, step by step

1. Choose a fixed tolerance $\epsilon = 0.005$.
2. Select a point $x = 1 + \frac{1}{n+1}$ that depends on $n$ and approaches the singularity at 1.
3. Use the integral test to establish the lower bound $\zeta(x) > n+1$.
4. Establish the upper bound for the partial sum $f_n(x) \le n$.
5. Calculate the difference $|\zeta(x) - f_n(x)| > (n+1) - n = 1$.
6. Observe that $1 > 0.005 = \epsilon$, proving that no single $N$ works for all $x > 1$.

Example

The video shows a graph where the gap between the limit function $\zeta(x)$ and the partial sums widens arbitrarily large as $x$ approaches 1 from the right, illustrating why uniform convergence fails on the entire domain $x>1$.

Common misconceptions

• Assuming that because the series converges pointwise for every $x>1$, it must also converge uniformly on the whole interval.
• Thinking that the counterexample requires a fixed $x$; in proving non-uniform convergence, $x$ is allowed to depend on $n$.

Watch the explanation

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