0

$$ (\lim_{n \to \infty} {\sum_{k = 1}^{n} {\|{x_k}\|} < \infty} ) \Longrightarrow (\lim_{n \to \infty} {\sum_{k = 1}^{n} {\|{x_k}\|}^2 < \infty} ) $$

chris
  • 3
  • 2
    $x_n = \frac{1}{\sqrt n}$ is an obvious counterexample. – MooS Feb 08 '15 at 12:20
  • so how does one prove that if a series is convergent, then the sum of the squares is convergent too? – chris Feb 08 '15 at 12:23
  • By using $|x_n|^2 \leq |x_n|$ for $n \gg 0$, provided $\sum |x_n|$ converges. Note that you need absolute convergence. – MooS Feb 08 '15 at 12:24
  • If I edit the question would you mind giving a complete proof? Please... – chris Feb 08 '15 at 12:26
  • 1
    $|x_n|^2 \leq |x_n|$ for $n \gg 0$ is actually the complete proof. You should now think about two things: Why does this inequality hold for $ n \gg 0$? Then check that this implies the assertion by comparison criteria. – MooS Feb 08 '15 at 12:30
  • So why does it hold? Because the sequence converges to zero, right? I can prove that. – chris Feb 08 '15 at 12:32

2 Answers2

1

If $\{a_n\}$ is a sequence of non-negative numbers such that $\sum_{n\ge 1}a_n$ converges, then $\lim_{n\to \infty}a_n=0\implies$ for an $\epsilon<1$ $\exists N,\ $ such that $\forall n\ge N$, $a_n\le \epsilon\implies \forall n\ge N$, $a_n^2\le \epsilon a_n$, then $\sum_{n\ge 1}a_n^2\le \sum_{n=1}^{N-1}a_n^2+\epsilon\sum_{n\ge N}a_n<\infty$

Samrat Mukhopadhyay
  • 16,277
1

$$ \lim_{n \to \infty} {\sum_{k = 1}^{n} {\|{x_k}\|} < \infty} \Longrightarrow \exists N.n \ge N \Rightarrow {\|{x_n}\|} \lt 1 $$ $$ {\|{x_n}\|} \lt 1 \Rightarrow {\|{x_n}\|}^2 \lt {\|{x_n}\|} $$ so by the comparison test $$ \lim_{n \to \infty} {\sum_{k = N}^{N+n} {\|{x_k}\|} < \infty} \Longrightarrow \lim_{n \to \infty} {\sum_{k = N}^{N+n} {\|{x_k}\|}^2 < \infty} $$ the finite set of terms with suffix $\lt N$ obviously has no bearing on the convergence of the infinite series

David Holden
  • 18,040