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Math Help - Nondecreasing Partial Sums

  1. #1
    Member RedBarchetta's Avatar
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    Nondecreasing Partial Sums

    Suppose that...

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

    <br />
s_1  \leqslant s_2  \leqslant s_3  \leqslant  \cdot  \cdot  \cdot  \leqslant s_n  \leqslant s_{n + 1}  \leqslant  \cdot  \cdot  \cdot <br />

    Would anyone mind explaining to me how <br />
s_{n + 1}  = s_n  + a_n <br />
is true?

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

    Let's write out the first couple terms:

    <br />
\sum\limits_{n = 1}^\infty  {n^2 }  = 1 + 4 + 9 +  \cdot  \cdot  \cdot  + n<br />

    So....

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

    What am I not getting?

    Thank you.
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  2. #2
    MHF Contributor
    Opalg's Avatar
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    Quote Originally Posted by RedBarchetta View Post
    Suppose that...

    <br />
\sum\limits_{n = 1}^\infty  {a_n } <br />
is an infinite series with <br />
a_n  \geqslant 0<br />
for all n. Then each partial sum is greater than or equal to its predecessor because <br />
s_{n + 1}  = s_n  + a_n <br />
    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).
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  3. #3
    Member RedBarchetta's Avatar
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    Thank you. Just a misprint in the book.
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