Growth of series with decreasing numerators and increasing deonimators

Question

It is known that $$H(n)=1+\dfrac12+\ldots+\dfrac1n$$ grows with the same rate as $\log n$.

Therefore, $$nH(n)=n\left(1+\dfrac12+\ldots+\dfrac1n\right)=\frac n1+\frac n2+\ldots+\frac nn$$ grows with the same rate as $n\log n$.

Now let $$B(n)=\dfrac{n}{1}+\dfrac{n-1}{2}+\ldots+\dfrac{1}{n}$$ Clearly $B(n)<nH(n)$, so $B(n)$ grows no faster than $\log n$. But exactly how fast does it grow?

Answer 1

Well

$$B(n) = \sum_{k=1}^n \frac{n+1-k}{k} = (n+1)H(n) - n \approx n\log n - n + \log n.$$

So it has the same asymptotic behaviour as $nH(n)$.

Answer 2

$$ B(n) = (n+1)H_n - n $$ ....