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IMC / 2001 / Problems / Day 1, P3

IMC 2001 · Day 1 · P3

hard

Find limt1(1t)n=1tn1+tn\lim\limits_{t \nearrow 1} (1-t) \sum\limits_{n=1}^{\infty} \dfrac{t^n}{1 + t^n}, where t1t \nearrow 1 means that tt approaches 1 from below.

Solution (official)

limt10(1t)n=1tn1+tn=limt101tlnt(lnt)n=1tn1+tn==limt10(lnt)n=111+enlnt=limh+0hn=111+enh=0dx1+ex=ln2.\begin{align*} \lim_{t \to 1-0} (1-t) \sum_{n=1}^{\infty} \frac{t^n}{1 + t^n} &= \lim_{t \to 1-0} \frac{1-t}{-\ln t} \cdot (-\ln t) \sum_{n=1}^{\infty} \frac{t^n}{1 + t^n} = \\ &= \lim_{t \to 1-0} (-\ln t) \sum_{n=1}^{\infty} \frac{1}{1 + e^{-n \ln t}} = \lim_{h \to +0} h \sum_{n=1}^{\infty} \frac{1}{1 + e^{nh}} = \int_0^{\infty} \frac{dx}{1 + e^x} = \ln 2. \end{align*}

How the field did

contestants scored
182
average (of 20)
5.62
solved (≥ 80%)
19.8%
near-0 (≤ 10%)
61.5%
discrimination
0.43

Score distribution (field cohort)

Computed on contestants with a meaningful total (field cohort); discrimination is the corrected item–total correlation.

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