Integral Test For Convergence

Applications

The harmonic series

$\sum_{n=1}^\infty \frac1n$

diverges because, using the natural logarithm, its derivative, and the fundamental theorem of calculus, we get

$\int_1^M\frac1x\,dx=\ln x\Bigr|_1^M=\ln M\to\infty \quad\text{for }M\to\infty.$

Contrary, the series

$\sum_{n=1}^\infty \frac1{n^{1+\varepsilon}}$

(cf. Riemann zeta function) converges for every ε > 0, because

$\int_1^M\frac1{x^{1+\varepsilon}}\,dx =-\frac1{\varepsilon x^\varepsilon}\biggr|_1^M= \frac1\varepsilon\Bigl(1-\frac1{M^\varepsilon}\Bigr) \le\frac1\varepsilon \quad\text{for all }M\ge1.$

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Integral Test For Convergence - Applications