Exercise 3.1

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Exercise 1: Show that if $\{s_n\}$ converges, so does $\{|s_n|\}$. Is the converse true?

ghostofgarborg · Accepted Answer

\def\eps{\varepsilon} Assume $s_n \to s$. Choose $\epsilon > 0$, and let $N$ be such that $|s_n - s| < \epsilon$ when $n > N$. Then $$ ||s_n|-|s|| \leq |s - s_n| < \epsilon $$ whenever $n > N$. This implies convergence of $|s_n|$. The converse is not true. Let $s_n = (-1)^n$. This sequence does not converge, even though $|s_n|$ does.

Exercise 3.1

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Exercise 3.1

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