Thread: Upper and Lower bound of the sequence log(n)

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

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!

Originally Posted by nmatthies1

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

Observe two facts.

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

$\displaystyle \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}} $

I see how to observe

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

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

$\displaystyle
\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}}
$

is true. Could you elaborate?

Let $\displaystyle F$ be an antiderivative for $\displaystyle f,$ thus $\displaystyle \int_{1}^{n-1}{f(t)\,dt}=F(n-1)-F(1)=\sum\limits_{k=2}^{n-1}{\big(F(k)-F(k-1)\big)}=\sum\limits_{k=2}^{n-1}{\int_{k-1}^{k}{f(t)\,dt}}.$

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