While proving some discrete Hardy-type inequalities I tried to prove the following limit for non-negative sequence $a(n)$ and $p>1$
$$\lim_{p\rightarrow1}\frac{1}{p-1}\left[\sum_{n=1}^{r}a(n)^{p}-\left(\sum_{n=1}^{r}a(n)\right)^{p}\right]=\sum_{n=1}^{r}a( n)\log \frac{a( n) }{\sum_{n=1}^{r}a( n)}$$
Since the limit is $\frac{0}{0}$, I tried to use Stolz–Cesàro Theorem but could not reach the result. I will be very appreciative of any help.
For convenience, it is not hard to note that the equality is preserved when multiplying all elements of $a(n)$ by a fixed positive $\lambda$; thus, we can prove the result for sequences such that $\sum_{n=1}^{r} a(n) = 1$, and the result will follow for all other sequences. (This doesn't handle the case that $a$ is the zero sequence, but I assume that is impossible, due to the $\sum_{n=1}^{r} a(n)$ on the denominator of a fraction. And if one defines the RHS by continuity for this case, then the equality holds anyway.)
Now, in fact, this can be tackled directly by L'Hôpital's. We have all the necessary conditions. $$ \lim_{p \to 1} \frac{\left(\sum_{n=1}^{r} a(n)^p\right) - 1^p}{p-1} = \lim_{p \to 1} \frac{\sum_{n=1}^{r} \log (a(n)) a(n)^p}{1} = \sum_{n=1}^{r} a(n)\log(a(n)). $$