Let bn = Show that In bn = Show that ln bn converges to ln x dx, and conclude th

Question

Let bn = Show that In bn = Show that ln bn converges to ln x dx, and conclude that 6n rightarrow e-1. (Rearranging terms, this shows that n! behaves like (n / e)n for large n. Look up "Stirling's Approximation" if you are interested in more details.)

Explanation / Answer

b(n) = (n!)^(1/n) v(n) = ln[(n)] = ln(n!)/n. So, b(n) = exp[v(n)]. v(n) = Sum[k=1,n, ln(k)/n] = (ln1 + ln2 + ... + ln(n) ) / n = Sum[k=1,n, ln(k/n)/n] + Sum[k=1,n, ln(n)/n] = = Sum[k=1,n, ln(k/n)/n] + ln(n) {Sum[k=1,n, b(n)] stands for the summation of the first 'n' terms of b(n)} Sum[k=1,n, ln(k/n)/n] tends to the integral of ln(x) between 0(+) and 1 (remember definition of Riemann's integral). This integral evaluates as follows: 1 I = S ln(x) dx = [xlnx -x] between 0 and 1. 0 [xlnx-x] = -1 when x=1... ...and tends to 0 when x-> 0(+). So, I = -1. ln(n) tends to +inf so, from 2, 3 and 4 comes: v(n) -> +inf. b(n) = exp[v(n)] so, b(n) also tends to +oo QED Let's put it more general... lim b(n) = lim n!^(1/n^p), where 'p' is a positive real number (n->+inf). v(n)= ln(b(n)) = ln n! / n^p Taking the same reasoning I used above, lim v(n) = lim (1/n^[p-1]) [ln n - Sum[k=1,n, ln(k/n)*1/n] As n->+inf, Sum[...] tends to the integral of ln(x) from 0(+) to 1 (which is -1). So, if... ...p=1 (previous case), then v(n)->+inf, so b(n) ->+inf ...p +inf, so b(n) -> +inf ...p>1, then v(n)-> 0, so b(n) -> 1 {for ln n/n^a -> 0 when a>0}

Related Questions

Navigate

• Browse (All)
• Browse L
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.