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.
Savionf
575
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
- 2114 views
- $2.00
Related Questions
- Write a Proof
- Find the arc length of $f(x)=x^{\frac{3}{2}}$ from $x=0$ to $x=1$.
- Volume of solid of revolution
- The cross sectional area of a rod has a radius that varies along its length according to the formula r = 2x. Find the total volume of the rod between x = 0 and x = 10 inches.
- 3 Multi-step response questions
- Help with Business Calculus problem.
- Solve only for the multiple choice part, the answer for the first box is 0
- Calculus Questions - Domains; Limits; Derivatives; Integrals