# Is that true ?

1. Jan 10, 2012

### 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 .

Last edited: Jan 10, 2012
2. Jan 11, 2012

### micromass

Staff Emeritus
Hello MIB!!

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

3. Jan 11, 2012

### MIB

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

4. Jan 11, 2012

### micromass

Staff Emeritus
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??

5. Jan 11, 2012

### MIB

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

Last edited: Jan 11, 2012
6. Jan 11, 2012

### MIB

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

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