A Disputable Problem

Calculus Level 3

Find the sum of the series

1 1 + 1 1 + 1 1 + 1 1 + . . . 1-1+1-1+1-1+1-1+...

1 2 \frac{1}{2} 1 None of these choices 0

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1 solution

If this sum converge to A then every subsequence would converge to same limit A, but in the sequence a n = i = 1 n ( 1 ) i + 1 a_n = \sum_{i=1}^n (-1)^{i+1} associated to the sum above you can take the subsequence with even terms, a 2 , a 4 , a 6 , . . . a_2 , a_4 , a_6, ... and lim n a 2 n \lim_{n \to \infty} a_{2n} = (1 - 1) + (1 - 1) + (1 -1) + ... = 0, and you can take the subsequence a 1 , a 3 , a 5 , . . . a_1 , a_3 , a_5 , ... and lim n a 2 n + 1 \lim_{n \to \infty} a_{2n + 1} = 1 + ( -1 +1) + (- 1 +1) + (- 1 +1) + (- 1 + 1) +... = 1 \Rightarrow the sum above doesn't converge.

Note: Euler believed wrongly that this sum precisely would converge to 0.5, it was necessary to aximomatize Aritmethic and Maths

Euler or Ramanujan?

Reynan Henry - 5 years, 5 months ago

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