How does the Euler constant cancel out when computing the limit of the area of the curvilinear trapezoid?

Conditions

• The asymptotic expansion $\sum_{i=1}^m \frac{1}{i} = \ln m + \gamma + \sigma_m$ is applied to both $m=2n$ and $m=n$.
• The limit is taken as $n \to \infty$.

Reasoning, step by step

1. Substitute the asymptotic expansion into the difference of harmonic series.
2. Expand the parentheses: $\ln 2n + \gamma + \sigma_{2n} - \ln n - \gamma - \sigma_n$.
3. Observe that $+\gamma$ and $-\gamma$ cancel each other.
4. Proceed to simplify the remaining logarithmic and infinitesimal terms.

Example

The video states: 'these two Euler constants cancel out exactly when we expand this parentheses'. An animation shows a red diagonal line striking out the two occurrences of $\gamma$.

Common misconceptions

• Thinking that $\gamma$ is zero or negligible from the start, rather than actively canceling out.
• Forgetting to distribute the negative sign to the second parenthesis, which would leave $+\gamma$ uncancelled.

Watch the explanation

Explore next

• How to simplify $\ln 2n - \ln n$ using logarithmic properties?
• What happens to the remainder terms $\sigma_{2n}$ and $\sigma_n$ in the limit?