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Question Number 132230 by pticantor last updated on 12/Feb/21

$$$$$$$$$$\mathit{S}=\underset{\mathit{k}=1}{\sum ^{+\mathrm{\infty }}}(\frac{1}{\sqrt{{\mathit{k}}^2-1}}-\frac{1}{\sqrt{{\mathit{k}}^2+1}})$$$$\mathit{S}=lor\mathit{S}=\mathrm{\infty }???$$

Answered by JDamian last updated on 12/Feb/21

$$\mathrm{\infty }because\frac{1}{\sqrt{k^2-1}}=\mathrm{\infty }k=1$$

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