use the integral test to determine whether the infinite series $sum_{n = 1}^{infty}\frac{13}{n^{2}+1}$ is…

use the integral test to determine whether the infinite series $sum_{n = 1}^{infty}\frac{13}{n^{2}+1}$ is convergent. (use symbolic notation and fractions where needed.) $int_{1}^{infty}\frac{13}{x^{2}+1}dx=$ determine whether the infinite series $sum_{n = 1}^{infty}\frac{13}{n^{2}+1}$ is convergent. the series converges. the series diverges.

use the integral test to determine whether the infinite series $sum_{n = 1}^{infty}\frac{13}{n^{2}+1}$ is convergent. (use symbolic notation and fractions where needed.) $int_{1}^{infty}\frac{13}{x^{2}+1}dx=$ determine whether the infinite series $sum_{n = 1}^{infty}\frac{13}{n^{2}+1}$ is convergent. the series converges. the series diverges.

Answer

Explanation:

Step1: Recall integral formula

We know that $\int\frac{1}{x^{2}+1}dx=\arctan(x)+C$. So, $\int_{1}^{\infty}\frac{13}{x^{2}+1}dx = 13\int_{1}^{\infty}\frac{1}{x^{2}+1}dx$.

Step2: Evaluate improper - integral

The improper - integral $\int_{1}^{\infty}\frac{1}{x^{2}+1}dx=\lim_{b\rightarrow\infty}\int_{1}^{b}\frac{1}{x^{2}+1}dx$. Since $\int\frac{1}{x^{2}+1}dx=\arctan(x)$, then $\lim_{b\rightarrow\infty}\int_{1}^{b}\frac{1}{x^{2}+1}dx=\lim_{b\rightarrow\infty}(\arctan(b)-\arctan(1))$.

Step3: Calculate the limit

We know that $\lim_{b\rightarrow\infty}\arctan(b)=\frac{\pi}{2}$ and $\arctan(1)=\frac{\pi}{4}$. So, $\lim_{b\rightarrow\infty}(\arctan(b)-\arctan(1))=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}$. Then $13\int_{1}^{\infty}\frac{1}{x^{2}+1}dx = 13\times\frac{\pi}{4}=\frac{13\pi}{4}$.

Answer:

$\frac{13\pi}{4}$, The series converges.