If $\lim_{n\to\infty}\frac{\ell_n}{\log n}=1$, does $n\log n\geq -C \iff (n+1)\ell_{n-1}\geq -C$ hold?

Question

Let $\ell_n:=\sum_{k=0}^{n}\frac{1}{k+1}.$

It is known that $\ell_n\sim\log n.$ i.e $\lim_{n\to\infty}\frac{\ell_n}{\log n}=1.$

Then, for $C\geq 0$ it is possible that

$n\log n\geq -C \iff (n+1)\ell_{n-1}\geq -C.$

Answer 1

$n$ is a real if we suppose $l_n= \sum_{0\le k \le n} \frac1{k+1}$

So $l_n = 0$ for $n <0$, $l_n = 1$ for $n \in [0,1[$, ...

Remarks:

1/ If $C\ge 0$, $-C \le 0$

2/ $ n \ln(n) >0$ if $n > 1$.

proof:

==> If $n> -1$, as $\ell_n \ge 0$ for all $n$, then $(n+1)\ell_{n \pm 1} \ge 0 \ge -C$.

<== if $(n+1)l_{n\pm 1} \ge 0$ , as $\ell_n \ge 0$ for all $n$, then $n+1 \ge 0$ and $n \ge -1$.

If we suppose $n\le 0$, then $\ell_{n\pm 1}\ge 0$ but $\ln(n)$ is not real ! and for $n \in ]0,1[$, the result is in function of $C$ as $x\ln(x) < 0$ and have a minimum in $x=1/\exp(1)$

conclusion: if $n \gt 0$, $\forall C \ge 1/\exp(1), \quad (n\log n\geq -C) \iff ((n+1)\ell_{n \pm 1}\geq -C)$.

but the conclusion is false if $n \le 0 $ as $n\ln(n)$ does not exist or if $C <1/\exp(1)$.