Rigorously Proving Direct Proportionality: Let's Find Out!

[MIB]

Is that true ?

Let be \Sigma_{n=1}^{\infty} a_n a series in ℝ .Suppose that \Sigma_{n=1}^{\infty} a_n is absolutely convergent . Suppose that for each Q \in N , \Sigma_{n=1}^{\infty} \frac{a_n}{Q^n} is convergent and \Sigma_{n=1}^{\infty} \frac{a_n}{Q^n} = 0.Then a_n = 0 for all n \in N.

My second question is : How Can I view direct proportionality rigorously without referring to the terms which are not precise like " Varying quantities " , " Variable " ? I tried doing this and I reached some ideas like making this variable as the value of a function. Is there a definitions all mathematicians work with different from that which says that y varies directly as x if there is a constant k , where y = kx ?, here we can't consider variable and constant as mathematical terms , because each element of set is a single element , but we use variable as a conventions only in writing as the value of a function for example in order to make writing easy , so Is there a definitions all mathematicians work with ?

Thanks .

 

[micromass]

Hello MIB!

Do you know some things about power series and analytic functions?

 

[MIB]

Hello MIB!

Do you know some things about power series and analytic functions?

yes , In fact I am using this to prove the uniqueness of the representations of functions as a power series , I yes I know about power series and series of functions and sequences of functions ... etc

 

[micromass]

yes , In fact I am using this to prove the uniqueness of the representations of functions as a power series , I yes I know about power series and series of functions and sequences of functions ... etc

Good. Do you know the theorem of "unique analytic continuation"?? That is: if X is a set with an accumulation point and if \sum{a_nx^n}=0 for all x\in X, then a_n=0??

Do you see what to take as X here??

 

[MIB]

Good. Do you know the theorem of "unique analytic continuation"?? That is: if X is a set with an accumulation point and if \sum{a_nx^n}=0 for all x\in X, then a_n=0??

Do you see what to take as X here??

no , I don't know it , but if I knew , I would put X as the set of 1/Q , where Q is a natural number and the number 0 , and the accumulation point is 0

 

[MIB]

ok must I begin know to prove the generalized theorem with this background in Mathematics ?