`sum_(n=1)^oo arctan(n)/(n^2+1)` Confirm that the Integral Test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series.
- print Print
- list Cite
Expert Answers
gsarora17
| Certified Educator
calendarEducator since 2015
write762 answers
starTop subjects are Math, Science, and Business
`sum_(n=1)^ooarctan(n)/(n^2+1)`
Integral test is applicable if f is positive, continuous and decreasing function on infinite interval `[k,oo)` where `k>=1` and `a_n=f(x)` . Then the series `sum_(n=1)^ooa_n` converges or diverges if and only if the improper integral `int_1^oof(x)dx` converges or diverges.
For the given series `a_n=arctan(n)/(n^2+1)`
Consider `f(x)=arctan(x)/(x^2+1)`
Refer the attached graph of the function. From the graph, we observe that the...
(The entire section contains 223 words.)
Unlock This Answer Now
Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.
Related Questions
- `sum_(n=1)^oo (-1)^n/sqrt(n)` Determine whether the series converges absolutely or...
- 1 Educator Answer
- Determine whether the series `sum_(1)^oo 0.6^(n-1) - 0.3^n`is convergent or divergent.
- 1 Educator Answer
- `sum_(n=1)^oo 1/(2n-1)` Use the Direct Comparison Test to determine the convergence or...
- 1 Educator Answer
- `1/3+1/5+1/7+1/9+1/11+...` Confirm that the Integral Test can be applied to the series. Then...
- 1 Educator Answer
- `sum_(n=1)^oo ln((n+1)/n)` Determine the convergence or divergence of the series.
- 1 Educator Answer