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??

  • 1
    $\begingroup$ Hint: $\frac1{2n-1}>\frac1{2n}$ $\endgroup$ – drhab Mar 4 '16 at 15:26
  • $\begingroup$ @drhab ah! So it must diverge $\endgroup$ – Allie Mar 4 '16 at 15:27
  • $\begingroup$ Yes, that is the correct conclusion. $\endgroup$ – drhab Mar 4 '16 at 15:27
  • 1
    $\begingroup$ 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 $\endgroup$ – Henry Mar 4 '16 at 15:28

For example:




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



  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 $v_n=1/n^2$

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