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Math Help - Upper and Lower bound of the sequence log(n)

  1. #1
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    Upper and Lower bound of the sequence log(n)

    Prove that for any integer n\geq 2
     \frac{1}{2}+\frac{1}{3}+ ... +\frac{1}{n} \leq log(n) \leq 1+ \frac{1}{2}+\frac{1}{3}+ ... +\frac{1}{n-1}.

    I am not sure where to start on this so maybe someone could give me a hint on what theorem(s) to use/where to start? thanks!
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  2. #2
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    Quote Originally Posted by nmatthies1 View Post
    Prove that for any integer n\geq 2
     \frac{1}{2}+\frac{1}{3}+ ... +\frac{1}{n} \leq log(n) \leq 1+ \frac{1}{2}+\frac{1}{3}+ ... +\frac{1}{n-1}.
    Observe two facts.

    \int_1^{n - 1} {\frac{1}{t}dt}  = \sum\limits_{k = 2}^{n - 1} {\left( {\int_{k - 1}^k {\frac{1}{t}dt} } \right)}

    \sum\limits_{k = 1}^{n - 1} {\frac{1}{{k + 1}}}  \leqslant \sum\limits_{k = 1}^{n - 1} {\left( {\int_k^{k + 1}{\frac{1}{t}dt} } \right)}  \leqslant \sum\limits_{k = 1}^{n - 1} {\frac{1}{k}}
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  3. #3
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    I see how to observe

    <br />
\int_1^{n - 1} {\frac{1}{t}dt} = \sum\limits_{k = 2}^{n - 1} {\left( {\int_{k - 1}^k {\frac{1}{t}dt} } \right)}<br />
.

    I don't quite see, however, how to observe that

    <br />
\sum\limits_{k = 1}^{n - 1} {\frac{1}{{k + 1}}} \leqslant \sum\limits_{k = 1}^{n - 1} {\left( {\int_k^{k + 1}{\frac{1}{t}dt} } \right)} \leqslant \sum\limits_{k = 1}^{n - 1} {\frac{1}{k}}<br />

    is true. Could you elaborate?
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  4. #4
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    Let F be an antiderivative for f, thus \int_{1}^{n-1}{f(t)\,dt}=F(n-1)-F(1)=\sum\limits_{k=2}^{n-1}{\big(F(k)-F(k-1)\big)}=\sum\limits_{k=2}^{n-1}{\int_{k-1}^{k}{f(t)\,dt}}.

    Now fix k\in\mathbb N so that k\le t\le k+1 and then \frac1{k+1}\le\frac1t\le\frac1k, so now integrate for k\le t\le k+1, thus \frac{1}{k+1}\le \int_{k}^{k+1}{\frac{dt}{t}}\le \frac{1}{k}\implies \sum\limits_{k=1}^{n-1}{\frac{1}{k+1}}\le \sum\limits_{k=1}^{n-1}{\int_{k}^{k+1}{\frac{dt}{t}}}\le \sum\limits_{k=1}^{n-1}{\frac{1}{k}}, now the rest is just a matter of make-up, finally \sum\limits_{k=1}^{n-1}{\frac{1}{k+1}}\le \ln n\le \sum\limits_{k=1}^{n-1}{\frac{1}{k}}, which is exactly what we wanted to prove.
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