Proving Divergence of \sum\frac{1}{2n+1}

[suluclac]

Prove that \(\displaystyle \sum\frac{1}{2n+1}\) diverges.
I understand that \(\displaystyle \sum\frac{1}{n}\) i.e. the harmonic series diverges (I say this because of the comparison test, that is, \(\displaystyle \frac{1}{2n+1}\leq\frac{1}{2n}\leq\frac{1}{n}\)).
However, this doesn't correctly imply that 1/(2n + 1) diverges.
Then I decided to use the integral test! WLOG, let the lower limit be even prime.
\(\displaystyle \int_2^\infty\frac{dx}{2x+1}=\frac{1}{2}\int_2^\infty\frac{2dx}{2x+1}=\frac{1}{2}[\ln{(2x+1)}]\Big|_2^\infty=\infty\)
Therefore, since the integral diverges, the series diverges.
Is this correct?

 

[Euge]

Hi suluclac,

Your work is correct, although one would technically have to justify (or just point out) that the sequence in question decreases to $0$ in order to apply the integral test.

Direct comparison test works here, too. For all $n \ge 2$, $2n + 1 \le 2n + (1/2)n = (5/2)n$. Therefore

$$\frac{1}{2n+1} \ge \frac{2/5}{n}\quad (n \ge 2)$$

Since the harmonic series diverges, so does $\sum \frac{2/5}{n}$; by direct comparison the series $\sum \frac{1}{2n+1}$ diverges.

 

[suluclac]

Although mathematically sufficient, it amazes me the fact that the inequality reverses as we multiply 1/n by 2/5!