2
$\begingroup$

Lets say that $$\sum_{n=a}^\infty f(n)$$ diverges. Does $$\sum_{n=a}^\infty \sqrt{f(n)}$$

necessarily diverge?

$\endgroup$
8
  • $\begingroup$ Do we have $f(n)>0$ for all $n$? $\endgroup$
    – ShakesBeer
    Jul 15, 2015 at 19:55
  • $\begingroup$ Not nessesarily $\endgroup$
    – user253055
    Jul 15, 2015 at 19:56
  • 2
    $\begingroup$ @AAron Then how can you write $\sqrt{f(n)}$ for $n \geq a$? $\endgroup$
    – Zhanxiong
    Jul 15, 2015 at 19:59
  • 3
    $\begingroup$ For the $f(n)\ge 0$ case (so we are working in the reals). if $f(n)$ does not have limit $0$, we are finished, and if $f(n)$ has limit $0$, Comparison. $\endgroup$ Jul 15, 2015 at 19:59
  • $\begingroup$ It would come out as an imaginary term then, right? $\endgroup$
    – user253055
    Jul 15, 2015 at 20:00

3 Answers 3

4
$\begingroup$

Suppose $\sum f(n)$ diverges. If $f(n)$ is unbounded or does not vanish, the result is immediate, so assume it is bounded and vanishing.

Observe that there exists some $k>0$ such that $f(n)<1$ for all $n>k$. This implies that $\sqrt{f(n)}>f(n)$ for any $n>k$. Since we know $\sum f(n)$ diverges, so must this series whose tail is composed of larger terms.

EDIT NOTE: This argument assumes that all terms stay in $\mathbb{R}$

$\endgroup$
2
  • $\begingroup$ in fact, you have assumed all the terms are positive. My answer in the second part says the same, but with the extra complication of removing that assumption $\endgroup$
    – ShakesBeer
    Jul 15, 2015 at 20:08
  • $\begingroup$ You are correct, but I was using "all" to encompass even those in $\sqrt{f(n)}$. I could have been more clear. $\endgroup$
    – Terra Hyde
    Jul 15, 2015 at 20:11
2
$\begingroup$

For the case when $f(n)\ge 0\forall n$, simply observe that $$\sum_{n\ge 0}f(n)\le \left(\sum_{n\ge 0}\sqrt{f(n)}\right)^2$$ which implies divergence of $\sum_{n\ge 0}\sqrt{f(n)}$ whence $\sum_{n\ge 0}f(n)$ diverges.

$\endgroup$
1
  • $\begingroup$ Concise and tidy $\endgroup$
    – Simon S
    Jul 15, 2015 at 21:41
1
$\begingroup$

There are two cases:

1) $f:\mathbb{N} \to \mathbb{C}$. Then the answer is no. Let $f(n)=1/n$ for $n$ odd and $f(n)=1/n-\epsilon i/n^2$. The sum of $f(n)$ obviously diverges but the sum of the square roots (being careful to use one branch of the root) will become, as $\epsilon \to 0$, $\sum \frac{(-1)^n}{\sqrt{n}}$ which is an alternating series.

2) $f:\mathbb{N} \to \mathbb{R}$. Then the answer is yes. The fact that $\sum f(n)$ diverges means that at least one of $A=\sum_{I_1}f(n)$ or $B=\sum_{I_2}f(n)$, where $A,B$ are the sums of the positive or negative terms of $f(n)$ respectively, diverge. WLOG $B$ diverges.

The sum $C=\sum_{I_2}\sqrt{f(n)}=\sum_{I_2}\sqrt{|f(n)|}i$ will then also diverge. Indeed, suppose it converges. Then eventually $|f(n)| \to 0$ for $n \in I_2$. Then $\sum_{I_2}f(n)<\sum_{I_2}\sqrt{|f(n)|}$ converges, contradiction (where in this sum it is implicit we have taken it over all sufficiently large $n$).

$\endgroup$
1
  • $\begingroup$ You assumed, apparently, the branch cut was taken along the positive real axis. It would benefit some readers as to this choice that renders the sign of the square root negative. One other issue pertains to the interchange of the limit on $\epsilon$ and the infinite series, although this really is not relevant to the thrust of the development herein. $\endgroup$
    – Mark Viola
    Jul 15, 2015 at 22:13

You must log in to answer this question.