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

Thanks in advance. Appreciate it.

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- Dec 2nd 2008, 03:36 PMthahachainaAnalysis (series)
Prove that if a series is convergent, then its partial sums, s_n, are bounded.

Thanks in advance. Appreciate it. - Dec 2nd 2008, 03:45 PMwatchmath
Well basically you need to show that every convergence sequence is bounded.

If $\displaystyle \{a_n\}$ convergence (say to$\displaystyle L$) then there is N such that for all $\displaystyle n\geq N$ we have

|$\displaystyle a_n-L|<1$.

Hence $\displaystyle |a_n|\leq |a_n -L|+L<L+1$ for $\displaystyle n\geq N$. Now $\displaystyle |a_n|\leq L+1+\sum_{i=1}^N |a_i|$ for every $\displaystyle n$, so it is bounded. - Dec 2nd 2008, 03:52 PMPlato
- Dec 2nd 2008, 04:51 PMMathstud28
A more direct way would warrant this. Let $\displaystyle S_N=\sum_{n=1}^{N}a_n$. So then for $\displaystyle \sum_{n=1}^{\infty}a_n$ to converge to a finite value, let's say $\displaystyle \xi$, we must have that $\displaystyle \lim_{N\to\infty}S_N=\xi$. But now assuming that we are talking about $\displaystyle \mathbb{R}^1$ we have that the defintion of a sequence converging to $\displaystyle \xi$ is for every $\displaystyle 0<\varepsilon$ there exists a $\displaystyle N$ such that $\displaystyle \forall{n\geqslant{N}}~~\left|S_n-\xi\right|<\varepsilon\implies{\xi-\varepsilon<S_n<\xi+\varepsilon}$. Now since $\displaystyle \xi,\varepsilon$ are real there exists a number let say $\displaystyle \gamma$ such that $\displaystyle \left|\varepsilon+\xi\right|<\gamma$. So it follows that $\displaystyle -\gamma<S_n<\gamma\implies\left|S_n\right|<\gamma$. Thus it is bounded.