Prove $\sum_{n=2}^{\infty}\frac{1}{n^2+e}<\frac{1}{2}$ Of course, you can use the following formula
$$\sum_{n = 1}^\infty \frac{1}{n^2 + a^2} = \frac{\pi\coth(\pi a)}{2a} - \frac{1}{2a^2},$$
but which is too "advanced". We want to find a solution only depending on inequality estimation only.
Maybe, we can obtain
\begin{align*}
\sum_{n=2}^{\infty} \frac{1}{n^2+e}&\le \sum_{n=2}^{100}\frac{1}{n^2+e}+\sum_{101}^{\infty}\frac{1}{n^2}=\sum_{n=2}^{100}\left(\frac{1}{n^2+e}-\frac{1}{n^2}\right)+\sum_{n=2}^{\infty}\frac{1}{n^2}\\
&=\frac{\pi^2}{6}-1+\sum_{n=2}^{100}\left(\frac{1}{n^2+e}-\frac{1}{n^2}\right)<\frac{1}{2},
\end{align*}
which is true by checking on machine, but too hard to compute by hand.
 A: I only use $\sum_{n=1}^\infty \frac{1}{n^k}$ for $k=2,4,6$.
Using $\frac{1}{n^2}-\frac{1}{n^2+e} = \frac{e}{n^2(n^2+e)}$, it suffices to show that $\sum_{n=2}^\infty \frac{1}{n^2(n^2+e)} > \frac{\pi^2-9}{9e}$. Then using $\frac{1}{n^4}-\frac{1}{n^2(n^2+e)} = \frac{e}{n^4(n^2+e)}$, it suffices to show $\sum_{n=2}^\infty \frac{1}{n^4(n^2+e)} < \frac{\pi^4-90}{90e}-\frac{\pi^2-9}{9e^2}$. Equivalently, we wish to show $\sum_{n=3}^\infty \frac{1}{n^4(n^2+e)} < \frac{\pi^4-90}{90e}-\frac{\pi^2-9}{9e^2}-\frac{1}{16(4+e)}$. So it suffices to show $\sum_{n=3}^\infty \frac{1}{n^6} < \frac{\pi^4-90}{90e}-\frac{\pi^2-9}{9e^2}-\frac{1}{16(4+e)}$, which is equivalent to $\frac{\pi^6}{945}-\frac{65}{64} < \frac{\pi^4-90}{90e}-\frac{\pi^2-9}{9e^2}-\frac{1}{16(4+e)}$, which is easy to do by hand/calculator (there's actually some room to spare).
A: This is a comment (I haven't enough reputation to comment).
Exact sum result:
$${{i\,\left(\psi_{0}(2-\sqrt{e}\,i)+\gamma\right)}\over{2\,\sqrt{e}
}}-{{i\,\left(\psi_{0}(\sqrt{e}\,i+2)+\gamma\right)}\over{2\,\sqrt{e
}}}$$
or
$$\frac{-3 e-1+(1+e) \pi  \sqrt{e} \coth \left(\sqrt{e} \pi \right)}{2 e (1+e)}$$
