- #1
courtrigrad
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Prove that if [tex] a_{n} > 0 [/tex] and [tex] \sum a_{n} [/tex] converges, then [tex] \sum a_{n}^{2} [/tex] also converges.
So if [tex] \sum a_{n} [/tex] converges, this means that [tex] \lim_{n\rightarrow \infty} a_{n} = 0 [/tex]. Ok, so from this part how do I get to this step: there exists an [tex] N [/tex] such that [tex]| a_{n} - 0 | < 1 [/tex] for all [tex] n > N \rightarrow 0\leq a_{n} < 1 [/tex]. Thus [tex] 0\leq a_{n}^{2} \leq a_{n} [/tex]. How did we choose [tex] |a_{n} - 0| < 1 [/tex]?
So if [tex] \sum a_{n} [/tex] converges, this means that [tex] \lim_{n\rightarrow \infty} a_{n} = 0 [/tex]. Ok, so from this part how do I get to this step: there exists an [tex] N [/tex] such that [tex]| a_{n} - 0 | < 1 [/tex] for all [tex] n > N \rightarrow 0\leq a_{n} < 1 [/tex]. Thus [tex] 0\leq a_{n}^{2} \leq a_{n} [/tex]. How did we choose [tex] |a_{n} - 0| < 1 [/tex]?