Question

Test the series for convergence or divergence. Say if convergence is conditional
∑1∞((−1)n+1n)÷(4n2+1)

Answer 1

\[\sum_1^{\infty}\frac{(-1)^n+n}{4n^2+1}\]?

Answer 2

oh i get it
\[\sum_1^{\infty}\frac{(-1)^{n+1}n}{4n^2+1}\]

Answer 3

maybe that one?

Answer 4

converges conditionally because it is alternating and the terms go to zero, so conditional for sure

Answer 5

\[\sum_{1}^{\infty}\left( \left( -1 \right)^{n +1}n \right)\div \left( 4n ^{2}+1 \right)\]

Answer 6

ok how do i show it tho?

Answer 7

well i am not sure what there is to show. the terms certainly go to zero because the degree of the denominator is bigger than the degree of the numerator. but it is not absolutely convergent

Answer 8

for conditional, as it is alternating, all you need is that the terms go to zero. i guess if you want to write something you can write
\[\lim_{n \rightarrow \infty} \frac{n}{4n^2+1}=0\]

Answer 9

by shear obviousness or l'hopital or whatever you like. but that just gives conditional convergence. absolute convergence would mean
\[\sum \frac{n}{4n^2+1}\] converges and it doesn't

Answer 10

ok cool thanks

Answer 11

the reason it doesn't is that the degree of the numerator is only one less that the degree of the denominator. that is how you do it with your eyeballs. the degree of the denominator would have to be greater than the degree of the numerator by MORE than one

Answer 12

if you want something mathematical looking to write, compare it to harmonic series
\[\sum \frac{1}{n}\] which you know does not converge. i.e. use "comparison test"