Solving a recurrence relation involving square roots
Dividing both sides by $n$ then introducing $S(m):=\frac{T(2^n)}{2^n}$ yields:

$$S(m)=S(m/2)+1+\frac{1}{m}$$

It follows that:
$$1&lt;S(m)-S(m/2)\leq 2$$
And further:
$$\forall k, 1&lt;S(m/2^k)-S(m/2^{k+1})\leq 2$$
Now summing for $k=1\dots \log(m)-1$ gives us:
$$S(m)=\Theta(\log(m))$$
And so:
$$T(n)=\Theta(n\cdot\log\log n)$$