Answer to Question #109705 in Real Analysis for Chinmoy Kumar Bera

Question

prove the following series 1-1/2^2-1/3^2+1/4^2-1/5^2-1/6^2+... is convergent

Answer 1

Let's show that a_n < 0, when n is big enough. Let us lead the expression of a_n to the common denominator. After the simplifications we obtain the following expression:

(-81n⁴ - 108n³ + 18n² + 72n - 23)/((3n + 1)²(3n+2)²(3n+3)²). Since the denominator is positive for every n=1,2,.. it is obvious that a_n < 0 when -81n⁴ - 108n³ + 18n² + 72n - 23 < 0(for example for n>10⁶ :) ). Now, we want to use the fact, that if a_n is asymptotically equivalent to the C/n^p , where p>1, C is some constant and a_n preserves the sign(so it is either >0 or <0 for the big enough n), then the respective series is convergent. Let's consider the simplificated expression for the a_n and divide the numerator and the denominator by the n⁴.

We obtain the following:

-(81+108/n - 18/n² - 72/n³ + 23/n⁴)/((3+1/n)²(3+2/n)²(3+3/n)²). This is asymptotically equivalent to the -81/(729(n+1)² )= -1/(9(n+1)²), which is asymptotically equivalent to the -1/n². Exactly what we needed to prove.