Bounds of the Harmonic Series

**p vs np** · September 9th 2008, 02:05 AM

Could anyone suggest a neat way to proving that the Harmonic Series,
Hn=1+1/2+1/3+1/4+...+1/n

is bounded by ln(n){lower bound} and ln(n)+1{upper bound} ?

ln is the natural logarithm.

**mr fantastic** · September 9th 2008, 04:22 AM

Originally Posted by p vs np

Could anyone suggest a neat way to proving that the Harmonic Series,
Hn=1+1/2+1/3+1/4+...+1/n

is bounded by ln(n){lower bound} and ln(n)+1{upper bound} ?

ln is the natural logarithm.

Read this: http://faculty.prairiestate.edu/skifowit/htdocs/sd1.pdf

**NonCommAlg** · September 9th 2008, 05:36 AM

Originally Posted by p vs np

Could anyone suggest a neat way to proving that the Harmonic Series,
Hn=1+1/2+1/3+1/4+...+1/n

is bounded by ln(n){lower bound} and ln(n)+1{upper bound} ?

ln is the natural logarithm.

it's clear for n = 1. for n > 1 write mean value theorem for $f(t)=\ln t, \ t > 1,$ on the interval $[x-1,x]$ to get: $\ln x - \ln (x-1)=\frac{1}{c},$ for some $x-1 < c < x.$ therefore:

$\frac{1}{x} < \ln x - \ln(x-1) < \frac{1}{x-1}.$ hence: $\sum_{k=2}^n \frac{1}{k} < \sum_{k=2}^n(\ln k-\ln(k-1)) < \sum_{k=2}^n \frac{1}{k-1},$ which gives us: $H_n - 1 < \ln n < H_n -\frac{1}{n}.$ thus: $\frac{1}{n}+\ln n < H_n < 1+\ln n,$ which

is a better result than what you asked!

**p vs np** · September 11th 2008, 06:21 AM

Compact and elegant.
Thanks

**Krizalid** · May 15th 2009, 12:00 PM

here's another way of proving those bounds:

for each $k\in\mathbb N$ is $\frac1{k+1}<\int_k^{k+1}\frac{dt}{t}<\frac1k.$

now it's $\frac1n+\ln n=\frac1n+\int_1^n\frac{dt}{t}=\frac1n+\sum_{k=1}^ {n-1}\int_k^{k+1}\frac{dt}t<\frac1n+\sum_{k=1}^{n-1}\frac1k=\sum_{k=1}^n\frac1k.$

on the other hand is $\sum_{k=1}^n\frac1k=1+\sum_{k=2}^n\frac1k=1+\sum_{ k=1}^{n-1}\frac1{k+1}<1+\sum_{k=1}^{n-1}\int_k^{k+1}\frac{dt}t=1+\int_1^n\frac{dt}t=1+\l n n.$

finally it's $\frac1n+\ln n<\sum_{k=1}^n\frac1k<1+\ln n.\quad\blacksquare$

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