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

• November 1st 2009, 12:17 PM
nmatthies1
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!
• November 1st 2009, 02:58 PM
Plato
Quote:

Originally Posted by nmatthies1
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.

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

$\sum\limits_{k = 1}^{n - 1} {\frac{1}{{k + 1}}} \leqslant \sum\limits_{k = 1}^{n - 1} {\left( {\int_k^{k + 1}{\frac{1}{t}dt} } \right)} \leqslant \sum\limits_{k = 1}^{n - 1} {\frac{1}{k}}$
• November 1st 2009, 05:18 PM
nmatthies1
I see how to observe

$
\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

$
\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?
• November 2nd 2009, 04:40 AM
Krizalid
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.