Convergence of $\sum\limits_{n=1}^{\infty}(-1)^n\frac{n+2}{n^2+n+1}$
Answer
First note that
\[\lim_{n \rightarrow \infty}\frac{\frac{n+2}{n^2+n+1}}{\frac{1}{n}}=1,\]
and hence by the Limit Comparision Test
\[\sum_{n=1}^{\infty}\frac{n+2}{n^2+n+1}\]
is divergent. Also since
\[\lim_{n\rightarrow}\frac{n+2}{n^2+n+1}=0,\]
by the Alternating Series Test
\[\sum_{n=1}^{\infty}(-1)^n\frac{n+2}{n^2+n+1}\]
converges conditionally but not absolutely.
574
The answer is accepted.
Join Matchmaticians Affiliate Marketing
Program to earn up to a 50% commission on every question that your affiliated users ask or answer.
- answered
- 3390 views
- $2.00
Related Questions
- < Derivative of a periodic function.
- Use Green’s theorem to compute $\int_C x^2 ydx − xy^2 dy$ where $C$ is the circle $x^2 + y ^2 = 4$ oriented counter-clockwise.
- Extremal values/asymptotes
- Is $\sum_{n=1}^{\infty}\frac{\arctan (n!)}{n^2}$ convergent or divergent?
- using maclaurin series for tan(x) and equation for length of cable to prove that x=
- Riemann Sums for computing $\int_0^3 x^3 dx$
- Applications of Integration [Calculus 1 and 2]
- Generalization of the Banach fixed point theorem