Best answer:
Let a(n) = (-1)^(n-1)*1/n and S(n) = [sum of i=1 to n] a(n) .
a(1) = 1 , S(1) = 1
a(2) = -1/2 , S(2) = 0.5
a(3) = 1/3 , S(3) = 0.833...
a(4) = -1/4 , S(4) = 0.58333...
a(5) = 1/5 , S(5) = 0.78333...
a(6) = -1/6 , S(6) = 0.61666...
a(7) = 1/7 , S(7) = 0.75952...
a(8) = -1/8 , S(8) = 0.63452...
and so on .
Let...
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Best answer: Let a(n) = (-1)^(n-1)*1/n and S(n) = [sum of i=1 to n] a(n) .
a(1) = 1 , S(1) = 1
a(2) = -1/2 , S(2) = 0.5
a(3) = 1/3 , S(3) = 0.833...
a(4) = -1/4 , S(4) = 0.58333...
a(5) = 1/5 , S(5) = 0.78333...
a(6) = -1/6 , S(6) = 0.61666...
a(7) = 1/7 , S(7) = 0.75952...
a(8) = -1/8 , S(8) = 0.63452...
and so on .
Let Osum(n) = S(2n-1) and Esum(n) = S(2n) .
If n increases then Osum(n) always decreases and Esum(n) always increases .
Osum(n) - Esum(n)
= S(2n-1) - S(n)
= S(2n-1) - [S(2n-1) + a(2n)]
= -a(2n)
= 1/(2n)
It is positive so always Osum(n) > Esum(n) .
And when n goes to +inf , Osum(n) - Esum(n) approaches 0 .
So they have the common convergent value when n goes to +inf .
Therefore , S(n) is convergent .
(The convergent value is ln2 ≒ 0.69314 , but this proof can not find it .)
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3 weeks ago