Bound on harmonic sum

Dependencies: (incomplete)

1. Bound on log

Let $H(n) = \sum_{i=1}^n \frac{1}{i}$. Then \[ \ln n + \gamma \le H(n) \le \ln(n+1) + \gamma \le \ln n + \gamma + \frac{1}{n} \]

Here $\gamma \approx 0.577216$ is the Euler-Mascheroni constant.

Proof

A proof of $\ln n + \gamma \le H(n) \le \ln(n+1) + \gamma$ can be found in a StackExchange answer.

Also, $\ln(n+1) - \ln n \le \ln \left( 1 + \frac{1}{n} \right) \le \frac{1}{n}$. Therefore, $\ln(n+1) \le \ln(n) + \frac{1}{n}$.

Dependency for: None

Info:

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• Number of transitive dependencies: 1

Transitive dependencies:

1. Bound on log