Bound for $\sum_{k=1}^\infty\left(\frac{1}{2^k+k^2}\right)$ I found for the series:
$$S=\sum_{k=1}^\infty\left(\dfrac{1}{2^k+k^2}\right)$$ a bound:
$$S\le\dfrac{\pi^2}{6+\pi^2}$$
which is in good agreement with the approximate value of $S$ calculated with Maple or Mathematica $(S=0.588239...)$, while the ratio: $\dfrac{\pi^2}{\pi^2+6}=0.621918...$ Is it possible to get a sharper bound for $S$? Thanks.

One idea:
$$\sum_{n=1}^{\infty} \frac{1}{n^2+2^n}=\sum_{n=1}^{k} \frac{1}{n^2+2^n}+\sum_{n=k+1}^{\infty} \frac{1}{n^2+2^n}$$
For some $k$
Now $\displaystyle \sum_{n=1}^{k} \frac{1}{n^2+2^n}$ is finite sum, so it's possible to calculate it directly.
In second series use inequality $\frac{1}{n^2+2^n} \leq \frac{1}{2^n}$, so:
$$\sum_{n=k+1}^{\infty} \frac{1}{n^2+2^n} \leq \sum_{n=k+1}^{\infty}\frac{1}{2^n}=\frac{1}{2^k} $$
You can get as good aproxximation as you want (for example for $k=10$ you get $4$-digit precision).
