數學問題: Sum to infinity

Since you have not stated the starting value of n, I would assume it to be 0. (If n should be starting from 1, you may just subtract 1 from my result)

Note that the given sum is not a G. P. since it involves the term cos(2nπ/3). In order to use the sum to infinity of a G. P., we have to do something.

Note the followings:
i) if n = 3m (a multiple of 3), cos(2nπ/3) = cos(6mπ/3) = cos(2mπ) = 1
ii) if n = 3m+1 (a multiple of 3 +1), cos(2nπ/3) = cos(6mπ/3 + 2π/3) = cos(2mπ + 2π/3) = cos(2π/3) = -1/2
iii) if n = 3m+2 (a multiple of 3 +2), cos(2nπ/3) = cos(6mπ/3 + 4π/3) = cos(2mπ + 4π/3) = cos(4π/3) = -1/2

If we want it to be a G. P., we need to rearrange the terms so that the multiples of 3 are grouped together, the multiples of 3 +1 are grouped together and the multiples of 3 +2 are grouped together.

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(If you have not learnt any analysis in University Mathematics, just skip this part and take it for granted that a rearrangement is possible in this question)

However, for rearrangement to be possible, the infinite sum must be absolutely convergent, meaning that the infinite sum of the absolute value of each term is convergent. But this can be easily proved because this sum is bounded above by a G. P. with common ratio 2/3.
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After a rearrangement, the sum splits into 3 sums:
Σ (n=0 to n=infinity) (-2/3)^n * cos(2nπ/3)
= Σ (m=0 to m=infinity) (-2/3)^(3m) * cos(2(3m)π/3) + Σ (m=0 to m=infinity) (-2/3)^(3m+1) * cos(2(3m+1)π/3) + Σ (m=0 to m=infinity) (-2/3)^(3m+2) * cos(2(3m+2)π/3)
= Σ (m=0 to m=infinity) (-2/3)^(3m) + Σ (m=0 to m=infinity) (-2/3)^(3m+1) * (-1/2) + Σ (m=0 to m=infinity) (-2/3)^(3m+2) * (-1/2) 【From what we have discussed in i), ii) and iii)】
= Σ (m=0 to m=infinity) (-2/3)^(3m) + (-1/2) Σ (m=0 to m=infinity) (-2/3)^(3m+1) + (-1/2) Σ (m=0 to m=infinity) (-2/3)^(3m+2)

Now, note that all the 3 sums are now G. P. with common ratio (-2/3)³=(-8/27) and differenent starting term.

For the first one, the starting term is 1 and common ratio is -8/27.
So the sum to infinity is 1 / [1-(-8/27)] = 1 / (35/27) = 27/35.

For the second one, the starting term is -2/3 and common ratio is -8/27.
So the sum to infinity is (-2/3) / [1-(-8/27)] = (-2/3) / (35/27) = -18/35

For the third one, the starting term is 4/9 and common ratio is -8/27.
So the sum to infinity is (4/9) / [1-(-8/27)] = (4/9) / (35/27) = 12/35

So the answer to the original question is
27/35 + (-1/2)(-18/35) + (-1/2)(12/35)
= 30/35
= 6/7

Sum to infinity(n->infinity) of (-2/3)^n * cos(2n丌/3)

=(-2/3)^0(1)+ (-2/3)(-1/2) + (-2/3)^2(-1/2) +(-2/3)^3(1) + (-2/3)^4(-1/2)+......+

=[ (-2/3)^0 + (-2/3)^3+ (-2/3)^6+...... ] + [ (-2/3)(-1/2) + (-2/3)^4(-1/2)+(-2/3)^7(-1/2)+.....] +[ (-2/3)^2(-1/2)+ (-2/3)^5(-1/2)+(-2/3)^8(-1/2)+......

=sum of a GP(with a=1 R=(-2/3)^3) - 1/2xsum of a GP(with a=-2/3 R=(-2/3)^3) - 1/2xsum of GP( with a=(2/3)^2 R= (-2/3)^3)

要返家吃飯，待會再答

2006-11-24 19:47:26 補充：
= 1/(1-(-2/3)^3) -1/2x ( (-2/3)/(1-(2/3)^3) -1/2x( (2/3)^2/(1-(-2/3)^3)=(27/35)x( 1 1/3 -4/27) =32/35

2006-11-24 19:49:06 補充：
對唔住，漏了＋號=(27/35)x( 1 1/3 -4/27) =32/35

2006-11-24 19:49:50 補充：
=(27/35)x( 1＋ 1/3 -4/27) =32/35

2006-11-24 19:54:20 補充：
對唔住，看錯了少許：應該是＝(27/35)(1 1/3 - 2/9)=6/7