Showing $\lim_{n\to\infty}\left\{\frac{n}{p-1}-\left[(\frac{n}{n+1})^p+(\frac{n}{n+2})^p+\cdots\right]\right\}=\frac12$ for $p>1$

For $ p > 1$, show that:

$$\lim_{n \to \infty} \left\{ \frac{n}{p-1} - \left[ \left(\frac{n}{n+1}\right)^p + \left(\frac{n}{n+2}\right)^p + \cdots\right] \right\}= \frac{1}{2}$$

I could compute the limit of the sum (up to the $n$-th term) divided by $n$ using the integral method:

$$\lim_{n \to \infty} \frac1n\sum_{k=1}^{n} \left(\frac{n}{n+k}\right)^p = \int_0^1 \frac{dx}{(1+x)^p} = \frac{1-2^{1-p}}{p-1}$$

Note: This is exercise 8, on page 22, from Mathematical Analysis: Functions, Limits, Series, Continued Fractions (ebook link via amazon.com), edited by L. A. Lyusternik and A. R. Yanpol'skii, first edition (1965).