Evaluating the sum <munder> <mo movablelimits="true" form="prefix">lim <mrow class="MJX-T

Santino Bautista

Santino Bautista

Answered question

2022-06-11

Evaluating the sum lim t k = t f ( k )

Answer & Explanation

lorienoldf7

lorienoldf7

Beginner2022-06-12Added 19 answers

If
| n = 1 a n | <
then for any ε > 0 there is an index N such that
| n = N a n | < ε .
The proof is actually very simple; let S N = n = 1 N 1 a n and let S = n = 1 a n . Then by definition, S N S as N , and therefore
| S S N | = | n = N a n |
must go to zero with N.

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