Morning.
I've written down some of my reasoning and arguments as to why the series diverges, however I am not certain I can safely conclude it diverges to $\infty$. Would you give it a look, please?
$$\sum_{n=1}^{\infty} \frac{1+\dfrac{1}{n}}{n} = \sum_{n=1}^{\infty} \frac{1}{n} + \sum_{n=1}^{\infty} \frac{1}{n^2}$$
Thus by recognising the $\zeta(n)$ when $n=1$ we know $\dfrac{1}{n}$ diverges,
as a sum of an infinite amount of $\dfrac{1}{2}$ results in $\infty$.
Hence, anything added to $\infty$ must also be $\infty$, assuming it is defined. So lets check if the last sum is defined.
By the p-test it is, and hence converges ($p>1$). So, divergence + convergence = divergence.
Could I however say $\infty + a|a\in\mathbb Q = \infty$, that is a given that it is any defined number.
And hence say that this sum diverges TO $\infty$?
There's no need to argue that $\infty + a = \infty$, since we can't literally perform "normal" arithmetic with "infinity", per se, though the idea behind it is sound. A divergent series grows without bound, so the sum of a divergent series and a convergent series is necessarily divergent, since adding a finite sum to an unbounded (infinite) sum is necessarily unbounded: it diverges to infinity.
So in the end, yes, you can certainly argue that $$\sum_{n=1}^{\infty} \frac{1+\dfrac{1}{n}}{n} = \left(\sum_{n=1}^{\infty} \frac{1}{n} + \sum_{n=1}^{\infty} \frac{1}{n^2}\right) = \infty$$