Nondecreasing Partial Sums

**RedBarchetta** · October 20th 2008, 11:33 PM

Suppose that...

$ \sum\limits_{n = 1}^\infty {a_n } $ is an infinite series with $ a_n \geqslant 0 $ for all n. Then each partial sum is greater than or equal to its predecessor because $ s_{n + 1} = s_n + a_n $

$ s_1 \leqslant s_2 \leqslant s_3 \leqslant \cdot \cdot \cdot \leqslant s_n \leqslant s_{n + 1} \leqslant \cdot \cdot \cdot $

Would anyone mind explaining to me how $ s_{n + 1} = s_n + a_n $ is true?

Let's say we have: $ \sum\limits_{n = 1}^\infty {n^2 } $ , how about letting n=2.

Let's write out the first couple terms:

$ \sum\limits_{n = 1}^\infty {n^2 } = 1 + 4 + 9 + \cdot \cdot \cdot + n $

So....

$ \begin{gathered} s_{n + 1} = s_n + a_n \hfill \\ s_3 = s_2 + a_2 \hfill \\ 14 \ne 5 + 4 \hfill \\ \end{gathered} $

What am I not getting?

Thank you.

**Opalg** · October 21st 2008, 12:45 AM

Originally Posted by RedBarchetta

Suppose that...

$ \sum\limits_{n = 1}^\infty {a_n } $ is an infinite series with $ a_n \geqslant 0 $ for all n. Then each partial sum is greater than or equal to its predecessor because $ s_{n + 1} = s_n + a_n $

That's not correct. It should be $s_{n + 1} = s_n + a_{n+1}$ ( $s_n$ is the sum of the first n terms; you need to add the (n+1)th term to that in order to get the sum of the first (n+1) terms).

**RedBarchetta** · October 21st 2008, 12:56 AM

Thank you. Just a misprint in the book.

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