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)

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

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