Upper and Lower bound of the sequence log(n)

**nmatthies1** · Nov 1st 2009, 12:17 PM

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!

**Plato** · Nov 1st 2009, 02:58 PM

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.

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

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

**nmatthies1** · Nov 1st 2009, 05:18 PM

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?

**Krizalid** · Nov 2nd 2009, 04:40 AM

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.

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

LinkBack

Thread Tools

Display

Upper and Lower bound of the sequence log(n)

Similar Math Help Forum Discussions

Upper Bound and Lower Bounds

Partial sets/ Least upper bound/Greastest lower bound/Lattices

Upper bound/Lower bound?

What is the upper & lower bound of 1.0 to one dec. place?

least upper bound and greatest lower bound

Search Tags