Does the following positive series converge? This seemed simple at first glance. I am trying to find out if the following series converges: $$\sum_1^\infty \frac{1}{2n +1} $$ I have tried various convergence tests outlined here: 
http://www.bates.edu/math-stat-workshop/files/2010/05/convergence_tests.pdf
 A: $$\infty=\frac {1}{2} \sum \frac {1}{n+1}=\sum \frac{1}{2n+2}\leq \sum \frac {1}{2n+1}$$
A: Note that if it converged, so would
$$\sum_{n\ge 1}\frac1{2n}=\frac12+\sum_{n\ge 1}\frac1{2n+2}\le\frac12+\sum_{n\ge 1}\frac1{2n+1}\;.$$
But then the harmonic series would converge:
$$\sum_{n\ge 1}\frac1n=\sum_{n\ge 1}\frac1{2n}+\sum_{n\ge 1}\frac1{2n+1}\;.$$
We know, however, that the harmonic series does not converge.
Alternatively, you can use the integral test with the function $f(x)=\frac1{2x+1}$:
$$\begin{align*}
\int_{x=1}^\infty\frac{dx}{2x+1}&=\lim_{a\to\infty}\int_1^a\frac{dx}{2x+1}\\
&=\lim_{a\to\infty}\left[\frac12\ln(2x+1)\right]_1^a\\
&=\frac12\lim_{a\to\infty}\ln(2a+1)-\frac12\ln 3\\
&=\infty\;.
\end{align*}$$
A: Since $\displaystyle\frac{1}{2n+1}\ge\frac{1}{3n}$ for $n\ge1$ and $\displaystyle\frac{1}{3}\sum_{n=1}^{\infty}\frac{1}{n}$ diverges (multiple of the harmonic series),
$\hspace{.3 in}\displaystyle\sum_{n=1}^{\infty}\frac{1}{2n+1}$ diverges by the Comparison Test.
A: Use equivalents: you have a series with positive terms, and
$$\dfrac1{2n+1}\sim_\infty\dfrac12\dfrac1n,$$
which diverges, hence it diverges.
