# Analysis (series)

• December 2nd 2008, 03:36 PM
thahachaina
Analysis (series)
Prove that if a series is convergent, then its partial sums, s_n, are bounded.

• December 2nd 2008, 03:45 PM
watchmath
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 for $n\geq N$. Now $|a_n|\leq L+1+\sum_{i=1}^N |a_i|$ for every $n$, so it is bounded.
• December 2nd 2008, 03:52 PM
Plato
Quote:

Originally Posted by thahachaina
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.
• December 2nd 2008, 04:51 PM
Mathstud28
Quote:

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

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. 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. Thus it is bounded.