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Corollary

Let $\sum a_n$ and $\sum b_n$ be series such that there is a number $N$ with $b_n \geq a_n \geq 0$ for $n>N$. Then if $\sum b_n$ converges, so does $ \sum a_n$

so far I think there are 3 cases:

case i: $b_n > a_n > 0$ case ii: $b_n = a_n >0$ case iii: $b_n = a_n = 0$

Pf

Assume $\sum a_n$ and $\sum b_n$ are series

$\forall \epsilon > 0$

case i:

Suppose $b_n > a_n > 0$

Then $\exists N \in \mathbb{N}$ $\forall n > N$ .

Consider $\sum b_n$ converges

Then $T_n = b_1 + .... + b_n$ is bounded

We have $S_n = a_1 +$ ... $+ a_n < b_1 +$ .... $+ b_n = T_n$

so $|b_n - a_n| < \epsilon$

Hence $\sum a_n$ converges

case ii & case iii trivial

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    $\begingroup$ so are you asking if your proof is correct? $\endgroup$ – DeepSea Mar 27 '16 at 2:18
  • $\begingroup$ yeah i feel like my ideas on the proof are scattered, is if sufficient? $\endgroup$ – David Mar 27 '16 at 2:19
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    $\begingroup$ It is sufficient, but I think using Cauchy criterion makes it easier. $\endgroup$ – Henricus V. Mar 27 '16 at 2:20
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We have: $0 \leq a_k \leq b_k\Rightarrow A_n = \displaystyle \sum_{k=N+1}^n a_k \leq \displaystyle \sum_{k=N+1}^n b_k= B_n\leq \displaystyle \lim_{n\to \infty} B_n=B$. Thus $\{A_n\}$ is an increasing sequence, and bounded above by $B$, thus converges. This means $\displaystyle \sum_{k=1}^\infty a_k$ converges.

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