morning observation

\sum_{n=1}^{\infty} \frac{1}{a^{n}} = \frac{1}{a-1}

What my morning’s observation is that if we consider something like:
\sum_{n=1}^{\infty} \frac{1}{2^{n}+n^{2}} \approx 0.5882390121819030646\ldots

\sum_{n=1}^{\infty} \frac{1}{3^{n}+n^{3}} \approx 0.3386248938590166285123\ldots

\sum_{n=1}^{\infty} \frac{1}{4^{n}+n^{4}} \approx 0.24096414317981403584949724\ldots

\sum_{n=1}^{\infty} \frac{1}{5^{n}+n^{5}} \approx 0.1877505985677791820123501\ldots

\sum_{n=1}^{\infty} \frac{1}{6^{n}+n^{6}} \approx 0.15415739781098121145677151253628\ldots

So,:
\sum_{n=1}^{\infty} \frac{1}{a^{n}+n^{a}} \propto 1/a

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