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Check if the following series is convergent or divergent:
$$\sum_{k=1}^{\infty}\frac{2}{\sqrt{4k^2+1}}$$ By the comparison test the series schould be divergent. I need a divergent series $\sum\limits_{k=1}^{\infty}b_k$ such that $0\le b_k \le \frac{2}{\sqrt{4k^2+1}}$ for all sufficient large natural k. But I don't know what $b_k$ could be, with $b_k=\frac{1}{k}$ I don't get $0\le b_k \le \frac{2}{\sqrt{4k^2+1}}$ for all sufficient large natural k. Could you help me?

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  • $\begingroup$ Try $b_k=\dfrac{2}{\sqrt{(2k+1)^2}}$, for all $k\in \mathbb N$. $\endgroup$ – Git Gud Apr 7 '15 at 21:25
  • $\begingroup$ Note that $\frac{2}{\sqrt{4k^2+1}}\gt \frac{2}{\sqrt{4k^2+12k^2}}$. $\endgroup$ – André Nicolas Apr 7 '15 at 21:33
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In short: it diverges because you can compare it with Harmonic series

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use that $$\frac{2}{\sqrt{4k^2+1}}\geq \frac{2}{2k+1}$$

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