Let bn = an− log n

Problem 11837

(American Mathematical Monthly, Vol.122, April 2015)

Proposed by I. Pinelis (USA).

Let a0 = 1, and for n ≥ 0 let an+1 = an + e^−an. Let bn = an− log n. For n ≥ 0, show that 0 < bn+1< bn and also show that limn→∞bn = 0.

Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.

We first show by induction that an≥ log(n + 1). It holds for n = 0 and for n ≥ 0,

an+1= f (an) ≥ f (log(n + 1)) = log(n + 1) + 1

n + 1≥ log(n + 1) + log

 1 + 1

n + 1



= log(n + 2)

where f (x) = x + e^−x is an increasing function for x ≥ 0.

Hence b_n≥ log(n + 1) − log(n) > 0, and

bn+1− bn= e^−an+ log

 1 − 1

n + 1



≤ −



− 1

n + 1− log

 1 − 1

n + 1



< 0

Since {bn}n≥0 is strictly decreasing positive sequence, it follows that it has a finite limit L ∈ [0, 1).

Note that

(n + 1)(bn+1− bn) = (n + 1) e^−bn n + log

 1 − 1

n + 1



→ e^−L− 1.

Finally, let Hn=Pn

k=11/k, then, by applying Stolz-Cesaro theorem, we have

0 = L

+∞ = lim

n→∞

bn

H_n

SC= lim

n→∞

bn+1− bn

1/(n + 1) = e^−L− 1.