Question

what test can i use to determine if the series converges or diverges?

Answer 1

\[\sum_{n=1}^{\infty}\frac{ \sqrt{n ^{4}+1} }{ n ^{3}+n ^{2}}\]

Answer 2

Comparison test. Compare to the series \(\sum \dfrac{1}{n}\). The comparison series diverges, so you'll have to show that
\[\frac{\sqrt{n^4+1}}{n^3+n^2}\ge\frac{1}{n}\]
to get the same result.

Answer 3

can i cross multiply?

Answer 4

Yes, you can manipulate the inequality in any way that works for you.

Answer 5

when i did that i got \[n \sqrt{n ^{4}+1} > n^{3}+n ^{2}\] what can i do next?

Answer 6

Divide both sides by n and then square both sides.

Answer 7

i got n^4+1>sqrt(n^2+n)

Answer 8

eyeball test

Answer 9

the numerator is essentially degree 2
the denominator is degree 3
3 is bigger than 2 only by 1
the degree of the denominator has to be more than one larger than the degree of the numerator

Answer 10

what test is that?

Answer 11

eyeball test

Answer 12

just look

Answer 13

if you want to please a teacher, use the limit comparison test
compare as a ratio with \(\frac{1}{n}\)

Answer 14

\[\lim_{n\to \infty}\frac{\frac{\sqrt{n^4+2}}{n^3+n^2}}{\frac{1}{n}}\]

\[=\lim_{n\to \infty}\frac{n\sqrt{n^4+2}}{n^3+n^2}=1\]

Answer 15

I support the eyeball test.

Answer 16

I think the eyeball test comes with practice though after doing more of these formally using comparison and limit comparison tests. It then becomes more obvious how to eyeball the series.