Show that $ \sum q^{-\deg \ p(x)} $ diverges

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Show that $\sum q^{-\deg \ p(x)}$ diverges, where the sum is over all monic irreducibles $p(x)$ in $K\left[x\right]$, where $k$ is finite field with $q$ elements.

First show that $\sum q^{-\deg \ f(x)}$ diverges and that $\sum q^{-2\deg \ p(x)}$ converges, where both sums over all monic polynomials $f(x)$ in $K[x]$. There are exactly $q^{n}$ monic polynomials of degree $n$ in $K\left[x\right]$.

Consider $\sum_{\deg \ f(x )\leq n} q^{-\deg \ f(x)}$, this sum is equal to $\sum_{m=1}^{n} q^{m}q^{-m}= n+1$, thus $\sum q^{-\deg \ f(x)}$ diverges. $ \ $ Similarly $\sum_{\deg \ f(x )\leq n} q^{-2\deg \ f(x)}$ is equal to $\sum_{m=1}^{n} q^{m}q^{-2m} \leq (1-\dfrac{1}{q})^{-1}$, thus $\sum q^{-2\deg \ f(x)}$ converges.

As would the rest of the proof?

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The number of irreducible polynomials of degree $n$ over $K$ is $$ \sum_{d|n} \mu(n/d) \,q^d \;, $$ so you have $$ \sum_{n=1}^N \frac{q^{-n}}{n} \left( \sum_{d|n} \mu(n/d) \,q^d \right) = \sum_{n=1}^N \frac{1}{n}\left(1-\sum_{p|n} q^{-n(p-1)/p} + \dots\right)$$ and all terms but the first converges as $n\rightarrow \infty$ and the divergence is proved. The second question is similar, but in this case you don't get the divergence of the harmonic series.