Odd series convergence Prove that we have following inequality:
$1+ \frac{1}{3} + \frac{1}{5} + ... + \frac{1}{397} > \frac{9}{4}$
Anybody can help me to figure it out?

A silly approach:
$$
\sum_{n=0}^{198} \frac{1}{2n+1} > \int_0^{198} \frac{dx}{2x+1} = \frac{1}{2} \log 397 > \frac{1}{2} \log 361 = \log 19,
$$
$$
e^{9/4} < 3^{9/4} < 3^{10/4} = 9 \sqrt{3} < 9 \cdot 2 = 18.
$$
A more sensible approach:
Write
$$
\sum_{n=1}^{199} \frac{1}{2n-1} > \sum_{n=1}^{\large 2^7} \frac{1}{2n-1} > \frac{1}{2} \sum_{n=1}^{\large 2^7} \frac{1}{n}
$$
then use Cauchy condensation.
