Answer is: option4
\[ \sum_{n=1}^{\infty}\frac{1}{n^2+1} \]Solution:
All of the series are continuous, decreasing, and positive valued.
(A) and (B) are divergent \(p\)-series with \(p \le 1\).
Using the integral test on (C), (D), and (E) we obtain:
\[ \int_{1}^{\infty}\frac{1}{2x+1}\,dx = \frac{1}{2}\ln(2x+1)\Big|_{1}^{\infty} = \infty \qquad \text{(C) is divergent} \]
\[ \int_{1}^{\infty}\frac{1}{x^2+1}\,dx = \arctan x\Big|_{1}^{\infty} = \frac{\pi}{4} \qquad \text{(D) is convergent} \]
\[ \int_{1}^{\infty}\frac{x}{x^2+1}\,dx = \frac{1}{2}\ln(x^2+1)\Big|_{1}^{\infty} = \infty \qquad \text{(E) is divergent} \]
