證明: (1) $\dps{\frac{\ln(1+x)}{x\ln 2}<\sum_{n=0}^\infty \frac{1}{2^n+x}<\frac{1}{1+x}+\frac{\ln(1+x)}{x\ln 2},\quad x>0}$;
(2) $\dps{\sum_{n=1}^\infty e^{-k^2x}<\frac{1}{2}\sqrt{\frac{\pi}{x}},\quad x>0.}$
證明: 由 $$\beex \bea \sum_{k=0}^n \frac{1}{2^k+x} &>\sum_{k=0}^n\int_k^{k+1}\frac{1}{2^t+x}\rd t\\ &=\int_0^{n+1}\frac{1}{2^t+x}\rd t,\\ \sum_{k=0}^n \frac{1}{2^k+x} &<\frac{1}{1+x}+\sum_{k=1}^n \int_{k-1}^k \frac{1}{2^t+x}\rd t\\ &=\frac{1}{1+x}+\int_0^n \frac{1}{2^t+x}\rd t \eea \eeex$$ 知 $$\bex \int_0^\infty\frac{1}{2^t+x}\rd t <\sum_{k=0}^n \frac{1}{2^k+x} <\frac{1}{1+x}+\int_0^\infty\frac{1}{2^t+x}\rd t. \eex$$ 又由 $$\beex \bea \int_0^\infty\frac{1}{2^t+x}\rd t &=\frac{1}{x\ln 2}\int_{1+x}^\infty \frac{1}{s}\cdot\frac{1}{s-x}\rd s\\ & =\frac{1}{x\ln 2}\int_{1+x}^\infty \sex{\frac{1}{s-x}-\frac{1}{s}}\rd s\\ & =\frac{1}{x\ln 2}\ln (1+x) \eea \eeex$$ 知結論成立.
由 $$\bex \sum_{k=1}^n e^{-k^2x} < \sum_{k=1}^n \int_{k-1}^k e^{-s^2x}\rd s=\int_0^n e^{-s^2x}\rd s \eex$$ 知 $$\bex \sum_{n=1}^\infty e^{-k^2x} <\int_0^\infty e^{-s^2x}\rd s =\frac{1}{2\sqrt{x}}\int_0^\infty t^{-\frac{1}{2}}e^{-t}\rd t =\frac{1}{2\sqrt{x}}\vGa\sex{\frac{1}{2}} =\frac{1}{2}\sqrt{\frac{\pi}{x}}. \eex$$
註記: 本題第 (2) 小題也可計算如下: $$\bex \int_0^\infty t^{-\frac{1}{2}}e^{-t}\rd t=\frac{1}{\sqrt{x}} \int_0^\infty e^{-t^2}\rd t =\frac{1}{\sqrt{x}}\cdot\frac{\sqrt{\pi}}{2}, \eex$$ 其中最後一步是因為 $$\bex \int_0^\infty e^{-t^2}\rd t =\frac{1}{2}\int_{-\infty}^{\infty}e^{-t^2}\rd t =\frac{1}{2}\sqrt{\iint_{\bbR^2}e^{-(s^2+t^2)}\rd s\rd t} =\frac{1}{2}\sqrt{\int_0^\infty e^{-r^2}\cdot 2\pi r\rd r} =\frac{1}{2}\sqrt{\pi}. \eex$$