document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1177273>Question 1177273</A>: <I><SPAN STYLE=\"word-break: break-all;\"> a(n) stands for sequence<BR>\n" );
document.write( "a(1)=2, a(2)=3,  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=a%28n%2B2%29=a%28n%2B1%29%2B1%2Fln%28a%28n%29%29 ALIGN=MIDDLE  ALT=\"a%28n%2B2%29=a%28n%2B1%29%2B1%2Fln%28a%28n%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "Prove that the limit <BR>\n" );
document.write( "when n->∞, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=a%28n%29ln%28a%28n%29%29%2Fn ALIGN=MIDDLE  ALT=\"a%28n%29ln%28a%28n%29%29%2Fn\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "exists and evaluate it </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #850492 by </B><A HREF=/tutors/aboutme.mpl?userid=CPhill><B>CPhill(1987)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CPhill><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=CPhill><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=850492\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let $a(1) = 2$ and $a(2) = 3$. We have the recurrence relation $a(n+2) = a(n+1) + \frac{1}{\log(a(n))}$.\r<BR>\n" );
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document.write( "First, we observe that $a(n)$ is an increasing sequence. Since $a(1) = 2 < 3 = a(2)$, we can show by induction that $a(n) < a(n+1)$ for all $n$.<BR>\n" );
document.write( "Assume $a(k) < a(k+1)$ for all $k \le n$. Then $a(n+1) > a(n)$, so $\log(a(n+1)) > \log(a(n))$, and $\frac{1}{\log(a(n))} > 0$.<BR>\n" );
document.write( "Thus, $a(n+2) = a(n+1) + \frac{1}{\log(a(n))} > a(n+1)$. Therefore, $a(n)$ is an increasing sequence.\r<BR>\n" );
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document.write( "Since $a(n)$ is increasing, $\log(a(n))$ is also increasing.<BR>\n" );
document.write( "Also, we can observe that $a(n) \to \infty$ as $n \to \infty$. If $a(n)$ were bounded, it would converge to some limit $L$. Then $L = L + \frac{1}{\log(L)}$, which implies $\frac{1}{\log(L)} = 0$, a contradiction. Thus, $a(n) \to \infty$.\r<BR>\n" );
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document.write( "Let's consider the differences:<BR>\n" );
document.write( "$a(n+2) - a(n+1) = \frac{1}{\log(a(n))}$<BR>\n" );
document.write( "Summing from $n=1$ to $N-2$, we get:<BR>\n" );
document.write( "$\sum_{n=1}^{N-2} (a(n+2) - a(n+1)) = \sum_{n=1}^{N-2} \frac{1}{\log(a(n))}$<BR>\n" );
document.write( "The left-hand side is a telescoping sum:<BR>\n" );
document.write( "$a(N) - a(2) = \sum_{n=1}^{N-2} \frac{1}{\log(a(n))}$<BR>\n" );
document.write( "$a(N) = 3 + \sum_{n=1}^{N-2} \frac{1}{\log(a(n))}$\r<BR>\n" );
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document.write( "Since $a(n) \to \infty$, $\log(a(n)) \to \infty$. Thus, $\frac{1}{\log(a(n))} \to 0$.<BR>\n" );
document.write( "The sum $\sum_{n=1}^{N-2} \frac{1}{\log(a(n))}$ grows slowly.\r<BR>\n" );
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document.write( "Let's consider $a(n+1) - a(n-1) = \frac{1}{\log(a(n-2))}$.<BR>\n" );
document.write( "We have $a(n+1) \approx a(n)$, so $a(n+1) - a(n) \approx 0$.<BR>\n" );
document.write( "Since $a(n) \to \infty$, $\log(a(n)) \to \infty$, so $\frac{1}{\log(a(n))} \to 0$.\r<BR>\n" );
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document.write( "We have $a(n+2) - a(n+1) = \frac{1}{\log(a(n))}$.<BR>\n" );
document.write( "Summing from $n=1$ to $N-1$, we get $a(N+1) - a(2) = \sum_{n=1}^{N-1} \frac{1}{\log(a(n))}$.<BR>\n" );
document.write( "Since $a(n)$ is increasing, we have $\frac{1}{\log(a(n))}$ decreasing.<BR>\n" );
document.write( "Thus, $\sum_{n=1}^{N-1} \frac{1}{\log(a(n))} \approx \int_1^{N-1} \frac{1}{\log(a(x))} dx$.<BR>\n" );
document.write( "Since $a(n+2) - a(n+1) = \frac{1}{\log(a(n))}$, we expect $a(n) \approx cn$ for some $c$.<BR>\n" );
document.write( "Then $\log(a(n)) \approx \log(cn) = \log(c) + \log(n)$.<BR>\n" );
document.write( "Thus, $a(N) \approx \sum_{n=1}^{N-2} \frac{1}{\log(cn)}$.\r<BR>\n" );
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document.write( "Let $a(n) \sim cn$. Then $cn+2c-cn-c = \frac{1}{\log(cn)}$.<BR>\n" );
document.write( "$c = \frac{1}{\log(cn)}$.<BR>\n" );
document.write( "$c \log(cn) = 1$.<BR>\n" );
document.write( "$c (\log(c) + \log(n)) = 1$.<BR>\n" );
document.write( "$c \log(n) \approx 1$.<BR>\n" );
document.write( "$c \approx \frac{1}{\log(n)}$.<BR>\n" );
document.write( "Then $a(n) \approx n/\log(n)$.\r<BR>\n" );
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document.write( "Let $a(n) \approx c n$. Then $c(n+2)-c(n+1)=\frac1{\log(cn)}$<BR>\n" );
document.write( "$c \approx \frac1{\log(cn)}$. Thus $c \log(cn) \approx 1$.<BR>\n" );
document.write( "$c(\log c + \log n) \approx 1$, so $c \log n \approx 1$. Thus $c\approx 1/\log n$.<BR>\n" );
document.write( "$a(n) \approx n/\log n$.<BR>\n" );
document.write( "Then $\frac{a(n) \log(a(n))}{n} \approx \frac{(n/\log n) \log(n/\log n)}{n} = \frac{\log n - \log\log n}{\log n} \to 1$.\r<BR>\n" );
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document.write( "Final Answer: The final answer is $\boxed{1}$<BR>\n" );
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