What if we rearrange the terms of a known series?
2 1 + 4 1 − 1 1 + 6 1 + 8 1 − 3 1 + 1 0 1 + 1 2 1 − 5 1 + ⋯
The series above is obtained by rearranging the terms of the alternating harmonic series. Determine the approximate value of the series rounded to three decimal places if it converges; otherwise, enter 0.
The answer is -0.347.
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2 solutions
Any reason why the following is wrong? 2 1 + 4 1 − 1 + 6 1 + 8 1 − 3 1 + 1 0 1 + 1 2 1 − 5 1 + . . . = 2 1 + 4 1 + 6 1 + . . . − ( 1 + 3 1 + 5 1 + 7 1 + . . . ) = 2 1 ( 1 1 + 2 1 + 3 1 + . . . ) − ( 1 + 3 1 + 5 1 + 7 1 + . . . ) = 2 1 ( 2 1 + 4 1 + 6 1 + . . . ) + 2 1 ( 1 + 3 1 + 5 1 + 7 1 + . . . ) − ( 1 + 3 1 + 5 1 + 7 1 + . . . ) = 2 1 ( 2 1 + 4 1 + 6 1 + . . . ) − 2 1 ( 1 + 3 1 + 5 1 + 7 1 + . . . ) = half of the orginal expression Then, using this logic, I got an answer of 0. Any reason why this is wrong?
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I think you should do the same thing, but working with partial sums. Notice that the alternating harmonic series is not absolutely convergent (it is conditionally convergent), so when you rearrange the terms of the series, you might change its sum. You can consider a partial sum and rearrange the terms in a convenient way, as I did in my solution. You can take a look of the wiki about absolute convergence.
It is a common rookie mistake / misconception made by those who are not familiar with calculus / analysis. As pointed out by Arturo, in order for us to justify rearranging terms, we have to ensure that sequence is absolutely convergent .
If the sequence is not absolutely convergent, then we can possibly rearrange the terms to get another limit, as you did above.
Very nice solution, Otto! I gave you my vote!
Ya.. ( + 1 ) ...same way ...... − 1 / 2 times expansion series of ln ( 1 + x ) at x = 1 .
PS:- I was going to write a one line solution for this problem but found you already here.. :-)
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My solution is a lengthy two lines, Comrade... go ahead and write your one-liner! ;)
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No..... My solution is essentially same as yours.... I would obviously skip some texts.... :-)
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@Rishabh Jain – Instead, go ahead and write a one line solution to this one ;)
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@Otto Bretscher – I would try my very best... !!
The series is convergent, Hence, the terms can be rearranged into − 1 + 2 1 − 3 1 + 4 1 + ⋯ = − ln ( 2 )
How come the same convergent series has two solutions. Particularly, when the brute force summation does converge to − ln ( 2 ) ??
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Your premise, "the series is convergent, hence, the terms can be rearranged", is flawed, as Mr. Presa explains in his comment. You need absolute convergence.
Let us use the notation H n = 1 + 2 1 + 3 1 + . . . + n 1 . The number H n is called the n t h - harmonic number. If we add the first 3 n first terms of the given series, we obtain the following expression S 3 n = k = 1 ∑ 2 n 2 k 1 − k = 1 ∑ n 2 k − 1 1 = k = 1 ∑ 2 n 2 k 1 − k = 1 ∑ n 2 k − 1 1 − k = 1 ∑ n 2 k 1 + k = 1 ∑ n 2 k 1 = k = 1 ∑ 2 n 2 k 1 − k = 1 ∑ 2 n k 1 + k = 1 ∑ n 2 k 1 . It is easy to see that S 3 n = − 2 1 H 2 n + 2 1 H n . Therefore, we obtain that S 3 n = − 2 1 ( H 2 n − ln 2 n ) + 2 1 ( H n − ln n ) − 2 1 ln 2 n + 2 1 ln n = − 2 1 ( H 2 n − ln 2 n ) + 2 1 ( H n − ln n ) + 2 1 ln 2 1 Now using that lim n → ∞ ( H n − ln n ) = γ , where γ is the Euler–Mascheroni constant, we obtain that lim n → ∞ S 3 n = 2 1 ln 2 1 ≈ − 0 . 3 4 7 .
There is one detail that is missing in my solution. Can somebody tell?
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You have to make sure that the sum of S 3 n + 1 , S 3 n + 2 approaches to that of S 3 n , which we can get from the squeeze theorem.
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Great! I agree with you. Otherwise, I could not be sure that our series is convergent. Thank you @Noam Pirani .
You can also see that the two subsequences S 3 n + 1 and S 3 n + 2 have the same limit as S 3 n , because lim n → ∞ ( S 3 n + 2 − S 3 n ) = 0 and lim n → ∞ ( S 3 n + 1 − S 3 n ) = 0 .
Combining each subtracted term with the term two places before, we can contract the series to − 2 1 + 4 1 − 6 1 + 8 1 − . . . = − 2 ln ( 2 ) ≈ − 0 . 3 4 7 , negative half of the alternating harmonic series.