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First of all I'm sorry about the wording of my title I'm not sure how to name it and my question was probably asked before but I'm just not sure how to word it properly

Either way my problem is that I know that the harmonic series $\sum_{n=1}^\infty \frac{1}{n}$ diverges. But how would I go on about to show that $\sum_{n=1}^\infty \frac{1}{k * n}$ where $k > 0$

I've tried the ratio test where I quickly realised it will not give me any conclusive solution. I also wasn't able to apply a direct comparison test.

Kind regards

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  • $\begingroup$ What does that mean? The number $n$ goes from $1$ to $\infty$. So, what's the meaning of $0<k<n$? $\endgroup$ – José Carlos Santos Apr 7 '18 at 7:16
  • $\begingroup$ I made a typo, sorry. Basically what I mean is that I know that a hyperharmonic series will converge. I'm looking for multiples of n, such as 2*n, 3*n etc. $\endgroup$ – genericCSstudent Apr 7 '18 at 7:20
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Note that $$\sum_{n=1}^N \frac{1}{k * n} = \frac {1}{k}\sum_{n=1}^N \frac{1}{ n}$$

Thus $$\sum_{n=1}^\infty \frac{1}{k * n} = \frac {1}{k}\sum_{n=1}^\infty \frac{1}{ n}$$

Since $\frac {1}{k}$ is a constant, the series will diverge.

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hint: $\sum \dfrac{1}{kn} = \dfrac{1}{k}\sum \dfrac{1}{n}$

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  • $\begingroup$ Holy! Of course.I feel stupid now. Thanks so much $\endgroup$ – genericCSstudent Apr 7 '18 at 7:24

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