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if $0≤a_n≤b_n$ then if $\sum a_n$ diverge so does $\sum b_n$

My approach :

Let $A_n = \sum_{1 \to k} a_k $ and $B_n = \sum_{1 \to k} b_k $

$A_n$ is an increasing sequence, so is $B_n$ We also have $A_n≤B_n$ for all n

Since $A_n$ diverges,$B_n$

I'm stuck here. Any help?

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  • $\begingroup$ This is Comparison Test, of which the proof is well known $\endgroup$ – Juniven Feb 19 '17 at 13:11
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Hint: As $A_n$ diverges to $\infty$ as increasing sequence, by definition for every $k \in \mathbb R$ exists $n \in \mathbb N$, such that $\forall m \geq n$ we have $A_m > k$. What does this imply for $B_m$?

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