Decide if the series converges and prove it using comparison test: $\sum_{k=1}^{\infty}\frac{1}{2k+3}$

$$\frac{1}{2k+3} <\frac{1}{2k}<\frac{1}{k}$$

and since $\sum_{k=1}^{\infty}\frac{1}{k}$ diverges (we have defined that in our readings, so I can claim it here), the complete series will diverge, too.

Did I do everything correctly? What makes me feel a bit unsure is that there is no "$\leq$" sign anywhere and when I look up the comparison test on the internet, I don't see any "<" or ">" in the definition.

Is it correct anyway?

  • 1
    $\begingroup$ By doing the chain of comparison as you did you prove that your sum will be less than $\sum_k 1/k$. This is fine, it is correct to say that $my\ sum < \sum_k 1/k$. Also we know that $\sum_k 1/k = \infty$ so your statement is saying $my\ sum < \infty$. However being less than infinity does not mean anything. $\infty -1$ is less than $\infty$ but still infinite. So saying something is less than infinity is equivalent to say that it can be whatever finite or can still be infinite. Therefore you can conclude that either you sum converges or diverges. No need to say this is a useless conclusion ! $\endgroup$ – Zubzub Aug 4 '16 at 12:40
  • $\begingroup$ @Zubzub what about $\sum_k (-1)^k < \sum_k 1$? It doesn't converge nor diverge. $\endgroup$ – cronos2 Aug 4 '16 at 13:13
  • $\begingroup$ @cronos2 Usually we say that a series that does not converge diverges. link. In your example the RHS = $\infty$ therefore the LHS can either converge or diverge. In this case it diverges. $\endgroup$ – Zubzub Aug 4 '16 at 13:22

You have to bound the series from below if you want to prove that it diverges. For example, $$ \sum_{k=1}^\infty\frac1{2k+3}>\sum_{k=1}^\infty\frac1{3k+3}=\frac13\sum_{k=1}^\infty\frac1{k+1}=\frac13\sum_{k=2}^\infty\frac1k=\infty. $$

  • $\begingroup$ I think I will never understand this stupid way of proofing convergence :( But thanks a lot for your answer! $\endgroup$ – cnmesr Aug 4 '16 at 12:35
  • $\begingroup$ @cnmesr Suppose that $a_n$, $b_n$ and $c_n$ are non-negative monotone real sequences such that $a_n\le b_n\le c_n$. If $a_n\to\infty$, then $b_n$ has to go to $\infty$ too. However, if $c_n\to\infty$, we cannot say anything about $b_n$. It might converge or diverge. Conversely, if $c_n$ converges, $b_n$ has to converge too. If $a_n$ converges, then we cannot say anything about $b_n$. $\endgroup$ – Cm7F7Bb Aug 4 '16 at 12:45

we can also say that $$\frac{1}{2k+3}>\frac{1}{3k}$$

  • 2
    $\begingroup$ this an answer to the question since the series $$\frac{1}{3}\sum_{k=1}^\infty\frac{1}{k}$$ is divergent $\endgroup$ – Dr. Sonnhard Graubner Aug 4 '16 at 13:05

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