∑k=1∞17k\sum\limits_{k = 1}^\infty {\frac{1}{{7k}}}k=1∑∞7k1
We apply the integral Cauchy criterion
∫1∞dx7x=17limt→∞lnx∣1t=17limt→∞(lnt−ln1)=17limt→∞ln=∞\int\limits_1^\infty {\frac{{dx}}{{7x}}} = \frac{1}{7}\mathop {\lim }\limits_{t \to \infty } \ln \left. x \right|_1^t = \frac{1}{7}\mathop {\lim }\limits_{t \to \infty } \left( {\ln t - \ln 1} \right) = \frac{1}{7}\mathop {\lim }\limits_{t \to \infty } \ln = \infty1∫∞7xdx=71t→∞limlnx∣1t=71t→∞lim(lnt−ln1)=71t→∞limln=∞
The series diverges since the corresponding improper integral diverges
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