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Suppose that $\sum a_n$ and $\sum b_n$ are series of positive terms and that there is a positive number $c$ s.t. $a_n \leq cb_n \forall n$. Show that if $\sum b_n$ converges, then $\sum a_n $ converges.

Is my proof sufficient?

Pf

Assume $\sum a_n$ and $\sum b_n$ are series of positive terms and that there is a positive number $c$ s.t. $a_n \leq cb_n \forall n$

Suppose $\sum cb_n$ converges.

Then $T_n = cb_1 + ... + cb_n = c(b_1 + ... + b_n)$.

We have, $S_n = a_1 + ..... + a_n \leq c(b_1 + ... +b_n) = T_n$

Thus, $S_n$ is bounded monotone sequence.

Therefore $\sum a_n$ converges

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    $\begingroup$ That's the right idea, but you haven't explicitly shown that all $S_n$ are bounded by the same value. $\endgroup$ – T.J. Gaffney Apr 14 '16 at 20:56
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It's correct but i would add a comment that $T_n$ is bounded.

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What you really need to show is that if $\sum b_n$ converges, then $\sum cb_n$ also converges.

Then, use the rest of your proof that compares the partial sums of $a_n$ against the partial sums of $cb_n$ with the inequality.

Now, you need to take a limit of both sides of the inequality to conclude.

At present, you are only saying something about the finite case, which is not enough -- it is the tail end of the sequence that matters, not earlier, possibly "transient behavior" of the sequence.

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