Does the square root of a sum of squares grow more slowly than the sum of terms?

Question

Suppose I have a non-negative sequence $a_n \geq 0$ satisfying $\sum a_n = \infty$. I’m trying to show that

$\dfrac{\displaystyle \sum a_n}{\sqrt{\displaystyle \sum a_n^2}}=\infty$

Is this always true? It seems to be true for $a_n = 1/n$ and $a_n = n$.

The fact that the numerator is larger than the denominator is due to the Cauchy-Schwarz inequality. — Adam Rubinson, Aug 09 '21 at 21:02

Greg Martin · Accepted Answer · 2021-08-09T23:07:38.937

9

No, this is false for divergent geometric series (other than $a_n=1^n$)—for example, $a_n=2^n$. A heuristic that leads to this type of example is that the sum of a finite geometric series is basically the same size as the largest term.

edited Aug 09 '21 at 23:07

answered Aug 09 '21 at 20:53

Greg Martin

78,820

+1: but II think you need to exclude the case of a geometric series with ratio $1$. – Rob Arthan Aug 09 '21 at 21:22

score -4 · Answer 2 · answered Aug 09 '21 at 21:09

-4

An easier way is to take $a_n = 1$ for all $n$. Then $\displaystyle \sum a_n = \sum 1 = \infty$ while $\dfrac{\displaystyle \sum a_n}{\sqrt{\displaystyle \sum a_n^2}}= 1\neq \infty$ .

answered Aug 09 '21 at 21:09

Wang YeFei

6,390

1

$\dfrac{\displaystyle \sum a_n}{\sqrt{\displaystyle \sum a_n^2}}= \lim_{n\to\infty}\frac{n}{\sqrt{n}}=\infty.$ – Adam Rubinson Aug 09 '21 at 21:14
1

This is incorrect - your example gives $\sqrt{n}$ which diverges. – Eric Aug 09 '21 at 21:16

Does the square root of a sum of squares grow more slowly than the sum of terms?

2 Answers2