Convergence/divergence using comparison test

Discuss the convergence or divergence of the series

$$1) \sum_{n=1}^\infty \frac{1}{2n}$$ $$2) \sum_{n=1}^\infty \frac{1}{2n-1}$$ $$3) \sum_{n=1}^\infty \frac{2}{n^2+3}$$

I have three following partial sums, and I wanted to discuss whether it converges or diverges using comparison test, that is

Suppose $$|a_n| \le b_n$$ for every $$n \sum_{n=1}^\infty b_n$$ converges, then $$\sum_{n=1}^\infty a_n$$ converges

Suppose $$0 \le b_n \le a_n$$ for every $$n \sum_{n=1}^\infty b_n$$ diverges, then $$\sum_{n=1}^\infty a_n$$ diverges

So by using this definition, I have come to the solution for 1) that

$$A) \sum_{n=1}^\infty \frac{1}{2n} = \frac12+\frac14+\frac16+....$$ where $$B) \sum_{n=1}^\infty \frac{1}{n} = 1+\frac12+\frac13+....$$

and since B is bigger than A, we say this partial sums diverge.

However, for 2) denominator is $$2n-1$$ and this gives me a headache.. also for 3)

Could I get some help solving these 2,3??

• Hint: $\frac1{2n-1}>\frac1{2n}$ – drhab Mar 4 '16 at 15:26
• @drhab ah! So it must diverge – Allie Mar 4 '16 at 15:27
• Yes, that is the correct conclusion. – drhab Mar 4 '16 at 15:27
• For 1) it would be better to say the terms in (A) are (greater than or) equal to half the terms in (B) and that (B) diverges – Henry Mar 4 '16 at 15:28

For example:

$$\frac1{2n-1}\ge\frac1{2n}$$

Also

$$\frac2{n^2+3}\le\frac2{n^2}$$

In both cases you have multiples of general terms of well know series: the former case a divergent one, and the latter a convergent.

HINT:

1. Use the fact that $\frac{1}{2n-1}>\frac{1}{2n}$ and then the fact that $\sum_n^\infty \frac{1}{n}$ diverges.
2. Use the fact that $\frac{2}{n^2+3}<\frac{2}{n^2}$ and then the fact that $\sum_n^\infty \frac{1}{n^2}$ converges (by p-series test or harmonic series).

Hint: Given series $\sum u_n$. Let $\sum v_n$ be auxiliary series s.t. $Lim_(n→\infty)$ $u_n/v_n =$a non zero finite number; then either both series converge or diverge.

1. Take $v_n=1/n$
2. Take $v_n=1/n$
3. Take $v_n=1/n^2$