- #1
estro
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Suppose a_n defined in the following way:
[tex]b_{2n}=a_{2n-1} [/tex]
[tex]b_{2n-1}=a_{2n} [/tex]
I know that [tex] \sum a_n [/tex] is convergent.
This how I proved that [tex] \sum b_n [/tex] is also convergent.
[tex]S_k=\sum_{n=1}^k b_n = \sum_{n=1}^k b_{2n} + \sum_{n=1}^k b_{2n-1} = \sum_{n=1}^k a_{2n-1} + \sum_{n=1}^k a_{2n} = \sum_{n=1}^k a_n \leq M [/tex]
Am I right?
[tex]b_{2n}=a_{2n-1} [/tex]
[tex]b_{2n-1}=a_{2n} [/tex]
I know that [tex] \sum a_n [/tex] is convergent.
This how I proved that [tex] \sum b_n [/tex] is also convergent.
[tex]S_k=\sum_{n=1}^k b_n = \sum_{n=1}^k b_{2n} + \sum_{n=1}^k b_{2n-1} = \sum_{n=1}^k a_{2n-1} + \sum_{n=1}^k a_{2n} = \sum_{n=1}^k a_n \leq M [/tex]
Am I right?