Is there any chance that this power series converges p-adically in the rational fieldin $mathbb{Q}$?












0












$begingroup$


Consider the p-adic power series $ sum _{n=1}^{infty} frac{x^n}{u^n}$. Consider the usual absolute value $|.|$ and p-adic absolute value $|.|_p$ on $ mathbb{Q}$, where $ u in mathbb{Q}$.



Is there any chance that this power series converges p-adically in the rational field $mathbb{Q}$ or $ sum frac{x^n}{u^n} in mathbb{Q} $ ?



We know that a power series $ sum a_n x^n$ converges p-adically iff $ |a_nx^n|_p to 0$.



Now for the given power series $ a_n=frac{1}{u^n}$ and $ |a_n|_p=|frac{1}{u^n}|_p to 0$ if $ p $ does not divide $ u$.



In the above case what would be the sum of the power series?



Can the sum be a rational number?



Help me






sequences-and-series p-adic-number-theory







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The OP surely means to ask whether the $p$-adic limit lies in $mathbb Q subseteq mathbb Q_p$?
    $endgroup$
    – Layer Cake
    Dec 28 '18 at 11:28








  • 1




    $begingroup$
    @LayerCake, Yes I mean that p-adic limit lies in $mathbb{Q}$
    $endgroup$
    – M. A. SARKAR
    Dec 28 '18 at 11:32






  • 3




    $begingroup$
    This obviously depends on $x,u$. For $x=p,u=1$ the series converges to $-1/(p-1)$
    $endgroup$
    – Wojowu
    Dec 28 '18 at 11:35






  • 6




    $begingroup$
    We are looking at a geometric series, which converges in $mathbb Q_p$ iff $|frac{x}{u}|_p<1$. The limit then is $1/(1-x/u)-1$ and whether it lies in $mathbb Q$ depends on $x$.
    $endgroup$
    – Layer Cake
    Dec 28 '18 at 11:39








  • 2




    $begingroup$
    Quite generally, the limit of a convergent geometric series $sum _{n=0}^{infty}q^n$ is $1/(1-q)$. You can prove this by looking at the n-th partial sum and then taking the limit (bearing in mind that $ q^n to 0$).
    $endgroup$
    – Layer Cake
    Dec 28 '18 at 11:46


















0












$begingroup$


Consider the p-adic power series $ sum _{n=1}^{infty} frac{x^n}{u^n}$. Consider the usual absolute value $|.|$ and p-adic absolute value $|.|_p$ on $ mathbb{Q}$, where $ u in mathbb{Q}$.



Is there any chance that this power series converges p-adically in the rational field $mathbb{Q}$ or $ sum frac{x^n}{u^n} in mathbb{Q} $ ?



We know that a power series $ sum a_n x^n$ converges p-adically iff $ |a_nx^n|_p to 0$.



Now for the given power series $ a_n=frac{1}{u^n}$ and $ |a_n|_p=|frac{1}{u^n}|_p to 0$ if $ p $ does not divide $ u$.



In the above case what would be the sum of the power series?



Can the sum be a rational number?



Help me






sequences-and-series p-adic-number-theory







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    The OP surely means to ask whether the $p$-adic limit lies in $mathbb Q subseteq mathbb Q_p$?
    $endgroup$
    – Layer Cake
    Dec 28 '18 at 11:28








  • 1




    $begingroup$
    @LayerCake, Yes I mean that p-adic limit lies in $mathbb{Q}$
    $endgroup$
    – M. A. SARKAR
    Dec 28 '18 at 11:32






  • 3




    $begingroup$
    This obviously depends on $x,u$. For $x=p,u=1$ the series converges to $-1/(p-1)$
    $endgroup$
    – Wojowu
    Dec 28 '18 at 11:35






  • 6




    $begingroup$
    We are looking at a geometric series, which converges in $mathbb Q_p$ iff $|frac{x}{u}|_p<1$. The limit then is $1/(1-x/u)-1$ and whether it lies in $mathbb Q$ depends on $x$.
    $endgroup$
    – Layer Cake
    Dec 28 '18 at 11:39








  • 2




    $begingroup$
    Quite generally, the limit of a convergent geometric series $sum _{n=0}^{infty}q^n$ is $1/(1-q)$. You can prove this by looking at the n-th partial sum and then taking the limit (bearing in mind that $ q^n to 0$).
    $endgroup$
    – Layer Cake
    Dec 28 '18 at 11:46
















0












0








0


2



$begingroup$


Consider the p-adic power series $ sum _{n=1}^{infty} frac{x^n}{u^n}$. Consider the usual absolute value $|.|$ and p-adic absolute value $|.|_p$ on $ mathbb{Q}$, where $ u in mathbb{Q}$.



Is there any chance that this power series converges p-adically in the rational field $mathbb{Q}$ or $ sum frac{x^n}{u^n} in mathbb{Q} $ ?



We know that a power series $ sum a_n x^n$ converges p-adically iff $ |a_nx^n|_p to 0$.



Now for the given power series $ a_n=frac{1}{u^n}$ and $ |a_n|_p=|frac{1}{u^n}|_p to 0$ if $ p $ does not divide $ u$.



In the above case what would be the sum of the power series?



Can the sum be a rational number?



Help me






sequences-and-series p-adic-number-theory







share|cite|improve this question











$endgroup$




Consider the p-adic power series $ sum _{n=1}^{infty} frac{x^n}{u^n}$. Consider the usual absolute value $|.|$ and p-adic absolute value $|.|_p$ on $ mathbb{Q}$, where $ u in mathbb{Q}$.



Is there any chance that this power series converges p-adically in the rational field $mathbb{Q}$ or $ sum frac{x^n}{u^n} in mathbb{Q} $ ?



We know that a power series $ sum a_n x^n$ converges p-adically iff $ |a_nx^n|_p to 0$.



Now for the given power series $ a_n=frac{1}{u^n}$ and $ |a_n|_p=|frac{1}{u^n}|_p to 0$ if $ p $ does not divide $ u$.



In the above case what would be the sum of the power series?



Can the sum be a rational number?



Help me






sequences-and-series p-adic-number-theory




sequences-and-series p-adic-number-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 28 '18 at 11:46







arifamath

















asked Dec 28 '18 at 11:02









arifamatharifamath

1196




1196








  • 2




    $begingroup$
    The OP surely means to ask whether the $p$-adic limit lies in $mathbb Q subseteq mathbb Q_p$?
    $endgroup$
    – Layer Cake
    Dec 28 '18 at 11:28








  • 1




    $begingroup$
    @LayerCake, Yes I mean that p-adic limit lies in $mathbb{Q}$
    $endgroup$
    – M. A. SARKAR
    Dec 28 '18 at 11:32






  • 3




    $begingroup$
    This obviously depends on $x,u$. For $x=p,u=1$ the series converges to $-1/(p-1)$
    $endgroup$
    – Wojowu
    Dec 28 '18 at 11:35






  • 6




    $begingroup$
    We are looking at a geometric series, which converges in $mathbb Q_p$ iff $|frac{x}{u}|_p<1$. The limit then is $1/(1-x/u)-1$ and whether it lies in $mathbb Q$ depends on $x$.
    $endgroup$
    – Layer Cake
    Dec 28 '18 at 11:39








  • 2




    $begingroup$
    Quite generally, the limit of a convergent geometric series $sum _{n=0}^{infty}q^n$ is $1/(1-q)$. You can prove this by looking at the n-th partial sum and then taking the limit (bearing in mind that $ q^n to 0$).
    $endgroup$
    – Layer Cake
    Dec 28 '18 at 11:46
















  • 2




    $begingroup$
    The OP surely means to ask whether the $p$-adic limit lies in $mathbb Q subseteq mathbb Q_p$?
    $endgroup$
    – Layer Cake
    Dec 28 '18 at 11:28








  • 1




    $begingroup$
    @LayerCake, Yes I mean that p-adic limit lies in $mathbb{Q}$
    $endgroup$
    – M. A. SARKAR
    Dec 28 '18 at 11:32






  • 3




    $begingroup$
    This obviously depends on $x,u$. For $x=p,u=1$ the series converges to $-1/(p-1)$
    $endgroup$
    – Wojowu
    Dec 28 '18 at 11:35






  • 6




    $begingroup$
    We are looking at a geometric series, which converges in $mathbb Q_p$ iff $|frac{x}{u}|_p<1$. The limit then is $1/(1-x/u)-1$ and whether it lies in $mathbb Q$ depends on $x$.
    $endgroup$
    – Layer Cake
    Dec 28 '18 at 11:39








  • 2




    $begingroup$
    Quite generally, the limit of a convergent geometric series $sum _{n=0}^{infty}q^n$ is $1/(1-q)$. You can prove this by looking at the n-th partial sum and then taking the limit (bearing in mind that $ q^n to 0$).
    $endgroup$
    – Layer Cake
    Dec 28 '18 at 11:46










2




2




$begingroup$
The OP surely means to ask whether the $p$-adic limit lies in $mathbb Q subseteq mathbb Q_p$?
$endgroup$
– Layer Cake
Dec 28 '18 at 11:28






$begingroup$
The OP surely means to ask whether the $p$-adic limit lies in $mathbb Q subseteq mathbb Q_p$?
$endgroup$
– Layer Cake
Dec 28 '18 at 11:28






1




1




$begingroup$
@LayerCake, Yes I mean that p-adic limit lies in $mathbb{Q}$
$endgroup$
– M. A. SARKAR
Dec 28 '18 at 11:32




$begingroup$
@LayerCake, Yes I mean that p-adic limit lies in $mathbb{Q}$
$endgroup$
– M. A. SARKAR
Dec 28 '18 at 11:32




3




3




$begingroup$
This obviously depends on $x,u$. For $x=p,u=1$ the series converges to $-1/(p-1)$
$endgroup$
– Wojowu
Dec 28 '18 at 11:35




$begingroup$
This obviously depends on $x,u$. For $x=p,u=1$ the series converges to $-1/(p-1)$
$endgroup$
– Wojowu
Dec 28 '18 at 11:35




6




6




$begingroup$
We are looking at a geometric series, which converges in $mathbb Q_p$ iff $|frac{x}{u}|_p<1$. The limit then is $1/(1-x/u)-1$ and whether it lies in $mathbb Q$ depends on $x$.
$endgroup$
– Layer Cake
Dec 28 '18 at 11:39






$begingroup$
We are looking at a geometric series, which converges in $mathbb Q_p$ iff $|frac{x}{u}|_p<1$. The limit then is $1/(1-x/u)-1$ and whether it lies in $mathbb Q$ depends on $x$.
$endgroup$
– Layer Cake
Dec 28 '18 at 11:39






2




2




$begingroup$
Quite generally, the limit of a convergent geometric series $sum _{n=0}^{infty}q^n$ is $1/(1-q)$. You can prove this by looking at the n-th partial sum and then taking the limit (bearing in mind that $ q^n to 0$).
$endgroup$
– Layer Cake
Dec 28 '18 at 11:46






$begingroup$
Quite generally, the limit of a convergent geometric series $sum _{n=0}^{infty}q^n$ is $1/(1-q)$. You can prove this by looking at the n-th partial sum and then taking the limit (bearing in mind that $ q^n to 0$).
$endgroup$
– Layer Cake
Dec 28 '18 at 11:46












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