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Let $$g(x)=\frac{xK_{n-1}(x)}{K_{n}(x)}$$ I want to prove that this function is increasing in $(0,+\infty)$ $\forall n=0,1,2\dots$. I tried to directly derive and using the recursion properties of $K_n$ but I had no luck. Thanks in advance for the help.

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It is shown in Theorem $6$ of this paper that for any $\nu\geq 0$, the function $$ (0,+\infty) \ni x \mapsto \frac{{xK'_\nu (x)}}{{K_\nu (x)}} $$ is strictly decreasing. Then for all $\nu\geq 0$, the function $$ (0,+\infty) \ni x \mapsto \frac{{xK_{\nu - 1} (x)}}{{K_\nu (x)}} \equiv - \nu - \frac{{xK'_\nu (x)}}{{K_\nu (x)}} $$ is strictly increasing.

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    $\begingroup$ Thanks a lot for the answer and reference. $\endgroup$ Commented Aug 12, 2022 at 15:50
  • $\begingroup$ It seems to me that this is valid also for negative $\nu$. Right? $\endgroup$ Commented Aug 15, 2022 at 10:00

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