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: 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.
A: Note that $\frac{1}{2k-1}+\frac{1}{2k+1}=\frac{4k}{4k^2-1}>\frac{4k}{4k^2}=\frac{1}{k}$.
Denote $1+\frac{1}{3}+\ldots+\frac{1}{2k-1}$ by $S_k$, then $S_{199}=1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\ldots+\frac{1}{397}$, so
\begin{align}
S_{199}&>1+\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\ldots\frac{1}{198}\\
&>1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\ldots+\underbrace{\frac{1}{128}+\ldots+\frac{1}{128}}_{32\;\text{terms}}+\frac{1}{130}+\ldots+\frac{1}{198} \\
&>1+\frac{1}{2}+\frac{1}{4}+\frac{2}{8}+\ldots+\frac{32}{128}\\
&=1+\frac{1}{2}+\underbrace{\frac{1}{4}+\frac{1}{4}+\ldots\frac{1}{4}}_{6\;\text{terms}} \\
&=1+\frac{1}{2}+\frac{6}{4} \\
&=3
\end{align}
This means $1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\ldots+\frac{1}{397}>3>\frac{9}{4}$.
