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.
 A: 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).
A: $(1/(2^k+k^2)) ≤ 1/2^k$. So replace the sum with this for $k>K$ with say $K=20$, and add the first K numbers manually. 
K = 4 gives an upper bound of 0.610907. K = 14 gives an upper bound of 0.588240. For 12 decimals, K = 26 gives an upper bound of 0.588,239,012,182. 
A: Looking at RIES for $S=0.588239012181903065$, there are some interesting numbers such as $$\frac{\sqrt{\pi}}{3}\approx 0.5908179503$$ $$\frac{1}{1+\log (2)}\approx 0.5906161091$$ $$\sqrt{\frac{\log (2)}{2}}\approx 0.5887050113$$ May I say that I really enjoy the last two !
By the way $$T=\sum_{k=1}^\infty\left(\dfrac{1}{3^k+k^3}\right) \approx 0.3386248939$$ while $$\frac{\frac{1}{5}+e}{\phi +7} \approx 0.3386250080$$ $$\frac{1}{(e-1)^2}\approx 0.3386968873$$
