Aha. Use a conditionally convergent, telescoping series: Instead of approaching the infinite sum via In[600]:= Sum[a[k], {k, 7}] Out[600]= a[1] + a[2] + a[3] + a[4] + a[5] + a[6] + a[7] In[601]:= Sum[a[k], {k, 9}] Out[601]= a[1] + a[2] + a[3] + a[4] + a[5] + a[6] + a[7] + a[8] + a[9] etc., take two odd terms for every even: In[596]:= Sum[a[2 k] + a[4 k - 3] + a[4 k - 1], {k, 3}] Out[596]= a[1] + a[2] + a[3] + a[4] + a[5] + a[6] + a[7] + a[9] + a[11] In[597]:= Sum[a[2 k] + a[4 k - 3] + a[4 k - 1], {k, 4}] Out[597]= a[1] + a[2] + a[3] + a[4] + a[5] + a[6] + a[7] + a[8] + a[9] + a[11] + a[13] + a[15] which, sooner or later, counts each term exactly once. Now put a(k) := (-1)^k/k - (-1)^(k+1)/(k+1) so that a[1], a[2], a[3], a[4], ... becomes -1-1^2/2, 1^2/2+1^3/3, -1^3/3-1^4/4, 1^4/4+1^5/5, . . . n terms of which clearly telescope to -1 + (-1)^n/(n+1), which clearly -> -1 as n-> oo. But the alternate, three-at-a-time summand simplifies to a[2 k] + a[-3 + 4 k] + a[-1 + 4 k] = 1/(1 - 4 k) + 1/(2 - 4 k) + 1/(3 - 4 k) + 1/(4 k) + 1/(1 + 2 k) This sum does *not* telescope, and according to Mathematica, converges to -1 - Log[2] ~ -1.69314718 < -1 (QED. This whole construction might be clearer with a[k] -> -a[k] := (-1)^(k+1)/(k+1) - (-1)^k/k .) --rwg On 2015-09-10 09:23, Eugene Salamin via math-fun wrote:
By suitable rearrangement, a conditionally convergent series can sum to any arbitrarily chosen value. Furthermore, a rearrangement exists such that the successive sums have any arbitrarily chosen limsup and liminf.
-- Gene
From: rwg <rwg@sdf.org> To: math-fun <math-fun@mailman.xmission.com> Sent: Wednesday, September 9, 2015 9:50 PM Subject: Re: [math-fun] Bogus proofs using rearrangement of conditionally convergent series
Mike Hirschhorn reminds me that there are infinitely many rearrangements of 1 - 1/2 + 1/3 - 1/4 + ... that sum to 42. --rwg
On 2015-09-09 19:19, rwg wrote:
You could just use Mathematica: For consecutive n, these three terms enumerate -(-)^n/n,
Out[567]= {1/(-3 + 4 n), 1/(-1 + 4 n), -1/(2 n)}
but taking two odds for every even:
In[568]:= Table[%, {n, 5}]
Out[568]= {{1, 1/3, -1/2}, {1/5, 1/7, -1/4}, {1/9, 1/11, -1/6}, {1/13, 1/15, -1/8}, {1/17, 1/19, -1/10}}
Adding them all up, In[569]:= Inactive[Sum][Total[%%], {n, \[Infinity]}]
Out[569]= Inactive[Sum][-1/(2*n) + 1/(-3 + 4*n) + 1/(-1 + 4*n), {n, Infinity}]
In[570]:= Activate[%]
Out[570]= Log[8]/2
I.e., 3/2 of the unpermuted series. --rwg
On 2015-09-09 10:03, James Propp wrote:
Can anyone provide a favorite web-reference?
For my upcoming Mathematical Enchantments column, I composed a proof that 1 - 1/2 + 1/3 - 1/4 + ... is both positive and zero, but it really doesn't fit into the (already over-long) article, so I'd like to outsource the job of showing "You can get nonsensical results by rearranging conditionally convergent series" to some website that already exists.
Jim Propp