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 $\displaystyle f(t)=\ln t, \ t > 1,$ on the interval $\displaystyle [x-1,x]$ to get: $\displaystyle \ln x - \ln (x-1)=\frac{1}{c},$ for some $\displaystyle x-1 < c < x.$ therefore:
$\displaystyle \frac{1}{x} < \ln x - \ln(x-1) < \frac{1}{x-1}.$ hence: $\displaystyle \sum_{k=2}^n \frac{1}{k} < \sum_{k=2}^n(\ln k-\ln(k-1)) < \sum_{k=2}^n \frac{1}{k-1},$ which gives us: $\displaystyle H_n - 1 < \ln n < H_n -\frac{1}{n}.$ thus: $\displaystyle \frac{1}{n}+\ln n < H_n < 1+\ln n,$ which
is a better result than what you asked!
here's another way of proving those bounds:
for each $\displaystyle k\in\mathbb N$ is $\displaystyle \frac1{k+1}<\int_k^{k+1}\frac{dt}{t}<\frac1k.$
now it's $\displaystyle \frac1n+\ln n=\frac1n+\int_1^n\frac{dt}{t}=\frac1n+\sum_{k=1}^ {n-1}\int_k^{k+1}\frac{dt}t<\frac1n+\sum_{k=1}^{n-1}\frac1k=\sum_{k=1}^n\frac1k.$
on the other hand is $\displaystyle \sum_{k=1}^n\frac1k=1+\sum_{k=2}^n\frac1k=1+\sum_{ k=1}^{n-1}\frac1{k+1}<1+\sum_{k=1}^{n-1}\int_k^{k+1}\frac{dt}t=1+\int_1^n\frac{dt}t=1+\l n n.$
finally it's $\displaystyle \frac1n+\ln n<\sum_{k=1}^n\frac1k<1+\ln n.\quad\blacksquare$