Question

show that if

Answer 1

\[\huge \sum _{k=1}^{\infty}a_k \text {converges and } \lim_{k \to \infty} \gamma_k=0\]
\[\huge \text{ where } (\gamma_k)_{k \ge 1}^{\infty} \text{is a sequence}\]
\[\huge \text{ then} \color {red}{\sum_{k=1}^\infty \gamma_k a_k} \text{converges}\]

Answer 2

This shouldn't be too hard... should it? lol
Let's see...

Answer 3

Oh wait, it is...

Answer 4

If only the convergent series was in fact, absolutely convergent :D

Answer 5

do you know the limit comparison test(qoutient test)

Answer 6

Of course. But I'm certain it only works for positive series...

Answer 7

that is if
\[\huge \sum a_k \text{ and }\sum b_k \text{ are positive,then } \lim _{k \to \infty}\frac{ a_k }{ b_k }=L\]
and \[\large \sum a_k \text{ converges and } \lim_{k \to \infty}\frac{ a_k }{ b_k }=0 \implies \sum b_k \text{converges}\]

Answer 8

sry i didnt mention that

Answer 9

You didn't say anything about these being positive...

Answer 10

Otherwise, it's as I said... it shouldn't be too hard...

Answer 11

oh i get it now we set
\[\Large a_k=a_k,b_k=\gamma_k a_k\]

Answer 12

Yeah

Answer 13

Wait, hold up there...

Answer 14

mistake the limit of gamma is infty

Answer 15

so \[\Large \lim_{k \to \infty}\frac{ 1 }{ \gamma_k }=0\]

Answer 16

No, wait...

Answer 17

gamma is a positive sequence, yes?

Answer 18

Google for Dirichlet test or Abel test

Answer 19

Why bother? If \(\large \gamma_k\) is a positive sequence that converges to zero, then it is bounded...
\[\Large \gamma_k \le U\] for some upper bound, U.

Then, since
\[\Large \sum_{k=1}^\infty a_k\] is a positive convergent series, then so is
\[\Large \sum_{k=1}^\infty Ua_k\]
And
\[\Large \gamma_k a_k \le Ua_k\] so after that, it's just direct comparison..