Series and sequences?

Question

sum((2n^2-1)/(n^2+1)) n=1 and goes to infinity...If it's convergent..find the sum

Answer 1

(2n^2-1)/(n^2+1), as n goes to infinity, approaches to 2. This you should be able to see first semester calculus, at worst, try dividing both the numerator and the denominator by n^2 or use L'Hospital's rule. Once you see that the terms you add up are not approaching to zero, you know that your series is diverging. However, just knowing that the terms in a series approaches to zero does not guarantee that a series converges, as the famous example of the series : "Sum (1/n) n=1 to infinity" illustrates.

Answer 2

i finished calculus II two month ago, but i forgot the divergent or con. tests. there are many tests but only one of them is usefull to solve the rational fuctions with degree two. i'm working on it.

but what do u think, which divergent test is usefull to solve this? if u are still there, u can email me

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i got it.

use the compereesion test.

for n>1

((2n^2-1)/(n^2+1)) > ((n^2+1)/(n^2+1)) because:
n^2 > 2 , n^2+1 = 2n^2-1 - n^2 + 2 > 2n^2-1 -2 + 2 =2n^2-1

so

((2n^2-1)/(n^2+1)) > ((n^2+1)/(n^2+1)) = 1 > ((n^2+1)/(2n^2+2)) = 1/2

1/2 + 1/2+ 1/2 +1/2 + .... is divergent so because ((2n^2-1)/(n^2+1)) > 1/2

((2n^2-1)/(n^2+1)) is also divergent.

for n=1

((2n^2-1)/(n^2+1)) = ((n^2+1)/(2n^2+2)) = 1/2

so for n=1, the series may diverges or converges but in this case we can assume it diverges.

as it was proved, for n>1 the series also diverges.

thanks, it is really bad to forget whole staff i learned before. ur question was a review for me.

Answer 3

sum from 1 to infinite of (2n^2-1) / (n^2+1)

The problem is that the formula gets larger the bigger n is. So it must not be convergent.

Answer 4

It's been a while, but I think it's divergent because the series' terms converge to 2, not zero?

Answer 5

I wish I could answer your math question but I took Algebra twice for eah of my 2 degrees.