A Triple Summation
Evaluate r = 3 ∑ ∞ n = 1 ∑ ∞ k = 1 ∑ ∞ r n + 1 ( k n )
The answer is 1.
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1 solution
I thin there is one problem .... In the first summation k k should be varied from 1 to n
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The binomial coefficients are zero if k > n .
@Mark Hennings I agree but statement made anything about aCb if b>a rather is meaningless than zero , I think. But as you say its zero I will take care of it.
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Note that ( b a ) = b ! ( a − b ) ! a ! = b ! a ( a − 1 ) ( a − 2 ) ⋯ ( a − b + 1 ) for integers a , b with 0 ≤ b ≤ a .
The expression on the RHS of this formula makes sense for any real number a and nonnegative integer b , and so we can use this to define the generalised Binomial coefficients ( b a ) for a ∈ R and b ∈ Z with b ≥ 0 . It is also clear, with this extended definition, that ( b a ) = 0 whenever a , b are integers with b > a ≥ 0 .
These extended Binomial coefficients are very useful. For example, we can now write ( 1 + x ) α = n = 0 ∑ ∞ ( n α ) x n ∣ x ∣ < 1 for any real number α . If α is a positive integer, the infinite series collapses to the usual finite form of the Binomial Theorem. Otherwise we get the correct coefficients of the Maclaurin series...
Using the Binomial Theorem, then summing GPs, then using the Method of Differences, r = 3 ∑ ∞ n = 1 ∑ ∞ k = 1 ∑ ∞ r n + 1 ( k n ) = r = 3 ∑ ∞ n = 1 ∑ ∞ r n + 1 2 n − 1 = r = 3 ∑ ∞ { 1 − r 2 r 2 2 − 1 − r 1 r 2 1 } = r = 3 ∑ ∞ ( r ( r − 2 ) 2 − r ( r − 1 ) 1 ) = r = 3 ∑ ∞ ( r − 1 ) ( r − 2 ) 1 = r = 1 ∑ ∞ r ( r + 1 ) 1 = r = 1 ∑ ∞ ( r 1 − r + 1 1 ) = 1