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Math Help - Analysis (series)

  1. #1
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    Analysis (series)

    Prove that if a series is convergent, then its partial sums, s_n, are bounded.

    Thanks in advance. Appreciate it.
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  2. #2
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    Well basically you need to show that every convergence sequence is bounded.
    If \{a_n\} convergence (say to  L) then there is N such that for all n\geq N we have
    | a_n-L|<1.
    Hence |a_n|\leq |a_n -L|+L<L+1 for n\geq N. Now |a_n|\leq L+1+\sum_{i=1}^N |a_i| for every n, so it is bounded.
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  3. #3
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    Quote Originally Posted by thahachaina View Post
    Prove that if a series is convergent, then its partial sums, s_n, are bounded.

    The statement that a series converges means that the sequence of its partial sums converges.
    Here is a standard theorem: If a sequence converges then it is bounded.
    Thus, simply apply that theorem.
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by thahachaina View Post
    Prove that if a series is convergent, then its partial sums, s_n, are bounded.

    Thanks in advance
    A more direct way would warrant this. Let S_N=\sum_{n=1}^{N}a_n. So then for \sum_{n=1}^{\infty}a_n to converge to a finite value, let's say \xi, we must have that \lim_{N\to\infty}S_N=\xi. But now assuming that we are talking about \mathbb{R}^1 we have that the defintion of a sequence converging to \xi is for every 0<\varepsilon there exists a N such that \forall{n\geqslant{N}}~~\left|S_n-\xi\right|<\varepsilon\implies{\xi-\varepsilon<S_n<\xi+\varepsilon}. Now since \xi,\varepsilon are real there exists a number let say \gamma such that \left|\varepsilon+\xi\right|<\gamma. So it follows that -\gamma<S_n<\gamma\implies\left|S_n\right|<\gamma. Thus it is bounded.
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