求极限 $\lim _{x \rightarrow+\infty} \sum_{n=1}^{\infty} \frac{x}{n^2+x^2}$.
【答案】 注意到
$$
\int_n^{n+1} \frac{x}{t^2+x^2} \mathrm{~d} t \leqslant \int_n^{n+1} \frac{x}{n^2+x^2} \mathrm{~d} t=\frac{x}{n^2+x^2}=\int_{n-1}^n \frac{x}{n^2+x^2} \mathrm{~d} t \leqslant \int_{n-1}^n \frac{x}{t^2+x^2} \mathrm{~d} t,
$$
所以有
$$
\sum_{n=1}^{\infty} \int_n^{n+1} \frac{x}{t^2+x^2} \mathrm{~d} t \leqslant \sum_{n=1}^{\infty} \frac{x}{n^2+x^2} \leqslant \sum_{n=1}^{\infty} \int_{n-1}^n \frac{x}{t^2+x^2} \mathrm{~d} t,
$$
又当 $x > 0$ 时,
$$
\begin{aligned}
&\sum_{n=1}^{\infty} \int_n^{n+1} \frac{x}{t^2+x^2} \mathrm{~d} t=\int_1^{+\infty} \frac{x}{t^2+x^2} \mathrm{~d} t=\left.\arctan \frac{t}{x}\right|_1 ^{+\infty}=\frac{\pi}{2}-\arctan \frac{1}{x} \rightarrow \frac{\pi}{2}(x \rightarrow+\infty), \\
&\sum_{n=1}^{\infty} \int_{n-1}^n \frac{x}{t^2+x^2} \mathrm{~d} t=\int_0^{+\infty} \frac{x}{t^2+x^2} \mathrm{~d} t=\left.\arctan \frac{t}{x}\right|_0 ^{+\infty}=\frac{\pi}{2},
\end{aligned}
$$
由极限夹逼准则得, $\lim _{x \rightarrow+\infty} \sum_{n=1}^{\infty} \frac{x}{n^2+x^2}=\frac{\pi}{2}$.


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