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sum(e - eta^^k): convergence or divergence?
#5
In the newsgroup news://sci.math I got a nice and concise answer by Prof. Israel.
At 16.08.2010 with the subject: "Series : divergent or convergent?" I gave t=e=exp(1), b=t^(1/t) and the notation for the residual-term r_k = t - b^^k as example.

This is the answer: (I inserted the correction of a wrong sign)

Code:
r_{k+1} = t - b^(b^^k) = t - b^(t - r_k)
        = t - t b^(-r_k)  (since t = b^t)
        = t (r_k ln(b) + O(r_k^2))  

By the ratio test, the series will converge if |t ln(b)| < 1.  
Since b^t = t says t ln(b) = ln(t), this is equivalent to 1/e < t < e.
Your case t=e is on the boundary of this, so we need another term.

r_{k+1} = t (r_k ln(b) - r_k^2 ln(b)^2/2 + O(r_k^3))
        = r_k - r_k^2/(2 e) + O(r_k^3)
            
This fits with r_k ~ 2e/k, which would indicate that the sum diverges.

[second post]
In fact, I believe we should have
     r_k ~= 2e/(k + ln(k)/(3 - 1/k))  as k -> infty.
Again, it diverges.

-- Robert Israel
    Department of Mathematics
    University of British Columbia Vancouver, BC, Canada

I think that solves the problem. @Henryk: Shall I still copy the problem into the TPID-section? (or the math-facts/does this still exist?)

Gottfried
Gottfried Helms, Kassel
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Messages In This Thread
RE: sum(e - eta^^k): convergence or divergence? - by Gottfried - 08/17/2010, 11:45 AM

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