what is the sum of (n-1)^2/(1+n^2) frin n=2 to n=infinity?

Question

I am asking wheter this converges or diverges. I know that the sequence converges to 1 but i cannot tell if the series does. I think it diverges because the sum of infinitely many one's is an large large number.

Answer 1

When you have a hunch that a series might diverge, you should apply the test for divergence: If you take the limit of the sequence and it doesn't equal to 0, then the series diverge.



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



divide the numerator and denominator by n^2 :



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



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



Taking the limit as n is approaching infinity, you get:



(1 - 0 + 0)/(0 + 1) = 1/1 = 1



Since the limit of the sequence is not zero, we definitely know the series doesn't converge.



Remark: You probably know this, but you can't apply this sort of test for convergence. If the limit of the sequence is 0, it doesn't mean that the sequence diverges. e.g., the series for 1/n: the sequence 1/n goes to zero as n approaches infinity, but we know that the infinite series for 1/n diverges.

Answer 2

you could chop up this sequence up into 2 infinite geometric sequence with easy ratios of two/5 and 3/5 as follows: ? [2^(n - a million) + 3^n]/5^(n + 2) (from n=a million to infinity) = ? [(a million/2)2^n + 3^n]/(25 * 5^n) (from n=a million to infinity), with the help of the guidelines of exponents = a million/25 ? [(a million/2)2^n + 3^n]/5^n (from n=a million to infinity), with the help of pulling out the 25 = a million/25 ? [(a million/2)2^n/5^n + 3^n/5^n] (from n=a million to infinity), with the help of splitting up the fraction = a million/25 ? [(a million/2)(2/5)^n + (3/5)^n] (from n=a million to infinity), with the help of the guidelines of exponents = a million/50 ? (2/5)^n (from n=a million to infinity) + a million/25 ? (3/5)^n (from n=a million to infinity). because of fact |2/5| < a million and |3/5| < a million, the two a style of infinite geometric sequence converges and so the sequence in question converges. --- As for the Ratio attempt, the shrink isn't a million if exact computed. With: a(n) = [2^(n - a million) + 3^n]/5^(n + 2) and a(n + a million) = [2^n + 3^(n + a million)]/5^(n + 3), we see that: a(n + a million)/a(n) = {[2^n + 3^(n + a million)]/5^(n + 3)}/{[2^(n - a million) + 3^n]/5^(n + 2)} = {5^(n + 2) * [2^n + 3^(n + a million)]}/{5^(n + 3) * [2^(n - a million) + 3^n]} = [2^n + 3^(n + a million)]/{5[2^(n - a million) + 3^n]}, because of fact 5^(n + 2)/5^(n + 3) = 5^(-a million) = a million/5 = (2^n + 3*3^n)/{5[(a million/2)2^n + 3^n]}, because of fact 3^(n + a million) = 3*3^n and a couple of^(n - a million) = (a million/2)2^n = [(2/3)^n + 3]/{5[(a million/2)(2/3)^n + a million]}, with the help of dividing numerator/denominator with the help of three^n. because of fact (2/3)^n --> 0 as n --> infinity, we see that as n --> infinity [(2/3)^n + 3]/{5[(a million/2)(2/3)^n + a million]} --> (0 + 3)/{5[(a million/2)(0) + a million]} = 3/5. So, |lim (n-->infinity) a(n + a million)/a(n)| = |3/5| = 3/5 < a million and the sequence converges with the help of the Ratio attempt. i'm hoping this facilitates!

Answer 3

(n-1)² / (n² + 1)

= (n² - 2n + 1) / (n² + 1)

= 1 - 2n / (n² + 1)



You're right, the sequence does converge to 1. And because it does, you'll be adding 1 + 1 + 1 + 1 + 1 + 1 + 1 + ....



In order for an infinite series to converge, it must converge to 0, not 1.

Answer 4

For an infinite series to converge T[n]→0 as n→∞



The series diverges

Answer 5

Integral Test

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

∫ dn -2 ∫ n dn / (n^2+1) from 2 to ∞ -----------(1)



Consider ∫ n dn / (n^2+1)

Let n^2+1 = u

when n=2, u=5

when n=∞, u =∞

2n dn = 2u

n dn = (1/2) du

The integral becomes

(1/2) ∫ du / u from 5 to ∞

(1/2) ln |u| from 5 to ∞

This part diverges

The entire integral (1) diverges

Answer 6

Each term gets succesively larger and approaches 1. So we are essentially adding 1 each time we get closer to infinity. Thus the sum is infinity.