Show that $\lim_{n \rightarrow \infty}\sum_{p=2}^na_{p}=\frac{3}{4}$

Question

For every natural number $p\geq2$ we have a real number $$a_{p}=\lim_{n \rightarrow \infty}\sum_{k=0}^n\frac{1}{p^{1+k}+p^{-k}+p+1}$$. Show that $$\lim_{n \rightarrow \infty}\sum_{p=2}^na_{p}=\frac{3}{4}$$
I think this problem is wrong because by induction we can se that $$\sum_{k=0}^n\frac{1}{p^{1+k}+p^{-k}+p+1}=\frac{p^{n+1}-1}{2(p-1)(p^{n+1}+1)}$$ so $a_{p}=\frac{1}{2(p-1)}$. So the final result will be the limit of $\frac{1}{2}H_{n-1}$ which is $\infty$. There is something messed up?