harder 3U question (1 Viewer)

apollo1 · Sep 29, 2011

can someone help me with part (ii) of this question:

tnx

hup · Sep 29, 2011

Drongoski · Sep 29, 2011

$ln(n!e^n) = lne^n + ln(n!) = n + ln (1\times2\times 3 \dots \times n)$ $= n + (ln1 + ln2+ \dots + ln (n) )$

$= (1+ln1 ) + (1+ln2) + (1+ln3) + \dots + (1+lnn)$

$= (ln e) + (ln 2e) + (ln 3e) + \dots + (ln (ne))$

$> e^{-1} + e^{-2} + e^{-3} + \dots + e^{-n}$

$= e^{-n} (1+e+e^2+e^3+ \dots + e^{n-1})$

$= e^{-n} \left ( \frac {e^n - 1}{e-1}\right )$

harder 3U question (1 Viewer)

apollo1

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