Answer to Question #91366 in Real Analysis for Sajid

Question #91366
Q. Choose the correct answer.
Q. The series ∑_(n=1)^∞▒〖(-1)〗^(n+1) n/(n^2+π) is
a. conditionally convergent for n>√π
b. absolutely convergent for n>π^2
c. divergent for n>0
d. none of the above
1
Expert's answer
2019-07-05T13:06:49-0400

1) The series of absolute values


n=1nn2+π\sum _{n=1}^{\infty } \frac{n}{n^2+\pi }


diverges because


nn2+πnn2+n2=121n,n2,\frac{n}{n^2+\pi } \ge \frac{n}{n^2+n^2 } = \frac{1}{2} \cdot \frac{1}{n}, n \ge 2,


and as you know the harmonic series diverges [https://en.wikipedia.org/wiki/Harmonic_series_(mathematics)].


2) The series


n=1(1)n+1nn2+π\sum _{n=1}^{\infty } \frac{(-1)^{n+1} n}{n^2+\pi }

converges by Leibniz's test [https://en.wikipedia.org/wiki/Alternating_series_test]. Absolute values decreases monotonically because the derivative is negative


(nn2+π)=πn2(n2+π)2<0,n2.\left(\frac{n}{n^2+\pi }\right)' = \frac{\pi -n^2}{\left(n^2+\pi \right)^2} < 0, n \ge 2.

3) nn is a local variable inside the series. Thus, the correct answer is automatically "d. none of the above". But we found that the original series converges conditionally (regardless of n).


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment