1. ## Nondecreasing Partial Sums

Suppose that...

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

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

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

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

Let's write out the first couple terms:

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

So....

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

What am I not getting?

Thank you.

2. Originally Posted by RedBarchetta
Suppose that...

$\displaystyle \sum\limits_{n = 1}^\infty {a_n }$ is an infinite series with $\displaystyle a_n \geqslant 0$ for all n. Then each partial sum is greater than or equal to its predecessor because $\displaystyle s_{n + 1} = s_n + a_n$
That's not correct. It should be $\displaystyle s_{n + 1} = s_n + a_{n+1}$ ($\displaystyle s_n$ 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).

3. Thank you. Just a misprint in the book.