- #1
drawar
- 132
- 0
Homework Statement
Let [itex](a_n)[/itex] be a sequence.
(i) Prove that if [itex]\sum\limits_{n = 1}^\infty {{a_n}} [/itex] converges, then [itex]\sum\limits_{n = 1}^\infty {\left( {{a_{2n - 1}} + {a_{2n}}} \right)} [/itex] also converges.
(ii) Prove that if [itex]\sum\limits_{n = 1}^\infty {\left( {{a_{2n - 1}} + {a_{2n}}} \right)} [/itex] converges and [itex]a_n \to 0[/itex], then [itex]\sum\limits_{n = 1}^\infty {{a_n}} [/itex] converges.
Homework Equations
The Attempt at a Solution
(i) Let [itex]{R_n} = \sum\limits_{k = 1}^n {{a_k}} [/itex], [itex]{S_n} = \sum\limits_{k = 1}^n {{a_{2k - 1}}} [/itex], and [itex]{T_n} = \sum\limits_{k = 1}^n {{a_{2k}}} [/itex]
Then [itex]{R_{2n}}={S_n}+{T_n}[/itex], since [itex]\sum\limits_{n = 1}^\infty {{a_n}} [/itex] converges, the sequence [itex](R_n)[/itex] converges, and so is the subsequence [itex](R_{2n})[/itex]. It follows that [itex]\sum\limits_{n = 1}^\infty {\left( {{a_{2n - 1}} + {a_{2n}}} \right)} [/itex].
(ii) Ok so I'm stuck on this part. I already have [itex](R_{2n})[/itex] converges and [itex](a_n)[/itex] is bounded, how can I go about proving that [itex](R_{n})[/itex] converges as well? Thank you!