Number theoretically geometric!
k = 1 ∑ ∞ ( 3 k − 2 k ) ( 3 k + 1 − 2 k + 1 ) 6 k = ?
The answer is 2.
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1 solution
Why can't you write the given series as \frac {3^k}{3^k-2^k} - \frac {3^(k+1)}{3^(k+1)-2^(k+1)} ? It seems working but in this case the series converges to 3! P.S. sorry for my latex :(
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You can do this, and the sum of the first K terms is then 3 − 2 3 − 3 K + 1 − 2 K + 1 3 K + 1 = 3 − 3 K + 1 − 2 K + 1 3 K + 1 but you then need to note that the second term on the RHS tends to − 1 as K → ∞ , so that the infinite sum is 3 − 1 = 2 .
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I understand your view but I thought the series was a telescopic series! In this case every term cancels out with the following, except the first 3/(3-2)=3 I'm sure I am wrong but I don't see why! (I find two solutions one with 3^k at the numerator and one with 2^k, so I am obviously wrong!) Thanks
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@Luca Ferrigno – Yes, it is a telescopic series. When we consider the finite sum k = 1 ∑ K ( 3 k − 2 k ) ( 3 k + 1 − 2 k + 1 ) 6 k = k = 1 ∑ K ( 3 k − 2 k 3 k − 3 k + 1 − 2 k + 1 3 k + 1 ) then every term, except the first, of the left-hand part of the bracket on the RHS cancels with a term from the right-hand part of the bracket on the RHS. Thus all terms cancel except for the first of the left-hand terms and the last of the right-hand terms. Thus this finite sum is equal to 3 − 2 3 − 3 K + 1 − 2 K + 1 3 K + 1 as I said. In my solution, the second term in my sum tended to 0 ; your second term tends to − 1 .
When summing a (simple) telescopic series, you generally end up with part of the first term, and part of the last. You can "throw away" the part from the last term when evaluating the infinite sum if that part tends to 0 as you include more terms (in the above notation, as K → ∞ ).
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@Mark Hennings – Oh! I always skip the step in which you consider the finite sum instead of the infinite one! Now I can correct my mistake! Thank you!!!
@Luca Ferrigno – Did the same
I did it by partial fractions involving telescopic sum. But yours is short one. Thanks for the solution.
Since ( 3 k − 2 k ) ( 3 k + 1 − 2 k + 1 ) 6 k = 3 k − 2 k 2 k − 3 k + 1 − 2 k + 1 2 k + 1 the series telescopes, and k = 1 ∑ K ( 3 k − 2 k ) ( 3 k + 1 − 2 k + 1 ) 6 k = 3 − 2 2 − 3 K + 1 − 2 K + 1 2 K + 1 = 2 − 3 K + 1 − 2 K + 1 2 K + 1 for all K ≥ 1 . Since the last term is O ( ( 3 2 ) K ) as K → ∞ , we deduce that the infinite sum is 2 .