S = ( 1 + 2 1 + 3 1 ) − ( 4 1 + 5 1 + 6 1 ) + ( 7 1 + 8 1 + 9 1 ) − ( 1 0 1 + 1 1 1 + 1 2 1 ) + …
If S can be expressed as:
C A B π D + F ln ( E )
for positive integers A , B , C , D , E , F where g cd ( A , C ) = 1 and B , E aren't any m t h power of a positive integer with m ∈ Z , m ≥ 2 .
Submit the value of A + B + C + D + E + F as your answer.
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Same method :) !!
By observing that after every 6 terms, the sign comes back, we can see
S = k = 0 ∑ ∞ 6 k + 1 1 + k = 0 ∑ ∞ 6 k + 2 1 + k = 0 ∑ ∞ 6 k + 3 1 − k = 0 ∑ ∞ 6 k + 4 1 − k = 0 ∑ ∞ 6 k + 5 1 − k = 0 ∑ ∞ 6 k + 6 1
A really simple way would be to use digamma function and I will do the easiest.
We know ψ ( s + 1 ) = ∫ 0 1 1 − x 1 − x s d x − γ
k = 0 ∑ ∞ 6 k + m + 1 1 = ∫ 0 1 1 − x 6 x m d x
k = 0 ∑ ∞ 6 k + 1 1 + k = 0 ∑ ∞ 6 k + 2 1 + k = 0 ∑ ∞ 6 k + 3 1 − k = 0 ∑ ∞ 6 k + 4 1 − k = 0 ∑ ∞ 6 k + 5 1 − k = 0 ∑ ∞ 6 k + 6 1 =
6 1 ( ∫ 0 1 1 − t t − 5 / 6 + 1 − t t − 4 / 6 + 1 − t t − 3 / 6 − 1 − t t − 2 / 6 − 1 − t t − 1 / 6 − 1 − t t − 0 / 6 d t )
= − 6 1 ( ψ ( 1 − 6 5 ) + ψ ( 1 − 6 4 ) + ψ ( 1 − 6 3 ) − ψ ( 1 − 6 2 ) − ψ ( 1 − 6 1 ) − ψ ( 1 − 6 0 ) )
= 6 1 ( ψ ( 1 ) + ψ ( 6 5 ) + ψ ( 6 4 ) − ψ ( 6 3 ) − ψ ( 6 2 ) − ψ ( 6 1 ) )
= 9 2 3 π + 3 ln ( 2 )
Riemann's Rearrangement theorem says that we cannot separate brackets and perform the calculation.
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@Kartik Sharma Hm... that's true, but either ways the answer to this question is still correct since it isn't necessary to split.
This is so tedious...
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Yeah! It's tedious!
Substituting cube roots of unity and eliminating also gives the same answer.
The sum is equal to S= x +y + z, where x = 1 - 1/4 + 1/7- 1/10 + 1/13 - ...=pi.sqrt(3)/9 + ln(2)/3 , y = 1/2 - 1/5 + 1/8 - 1/11 + ... = pi.sqrt(3)/9 - ln(2)/3 , z = 1/3 - 1/6 +1/9 - 1/12 + ... = ln(2)/3. So S = 2.pi.sqrt(3)/9 + ln(2)/3.
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Wait it's not tedious :
S can be written as :
S = r = 0 ∑ ∞ 3 r + 1 ( − 1 ) r + r = 0 ∑ ∞ 3 r + 2 ( − 1 ) r + r = 0 ∑ ∞ 3 r + 2 ( − 1 ) r
⇒ S = r = 0 ∑ ∞ ( − 1 ) r ∫ 0 1 x 3 r d x + r = 0 ∑ ∞ ( − 1 ) r ∫ 0 1 x 3 r + 1 d x + r = 0 ∑ ∞ ( − 1 ) r ∫ 0 1 x 3 r + 2 d x
Interchanging summation and integral we have :
S = ∫ 0 1 ( 1 + x + x 2 ) r = 0 ∑ ∞ ( − 1 ) r ( x 3 r ) d x
⇒ S = ∫ 0 1 1 + x 3 ( 1 + x + x 2 ) d x
Doing this is easy :
S = ∫ 0 1 1 + x 3 x 2 d x + ∫ 0 1 x 2 − x + 1 d x
These integrals are quite trivial and hence we have the answer as :
S = ( 3 1 ln ( x 3 + 1 ) + 3 2 arctan ( 3 2 x − 1 ) ) ∣ 0 1 = 9 2 3 π + 3 ln ( 2 )