# 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: $\frac1{2n-1}>\frac1{2n}$ Mar 4, 2016 at 15:26
• @drhab ah! So it must diverge Mar 4, 2016 at 15:27
• Yes, that is the correct conclusion. Mar 4, 2016 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 Mar 4, 2016 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 $v_n=1/n^2$