Proving that $limlimits_{xto 1^{-}}frac{1}{ln(1-x)}sumlimits_{n=0}^{infty}x^{b^n}=-frac{1}{ln(b)}$












8












$begingroup$


I conjecture that :
$$forall binmathbb{N}setminuslbrace0,1rbrace,limlimits_{xto 1^{-}}frac{1}{ln(1-x)}sumlimits_{n=0}^{infty}x^{b^n}=-frac{1}{ln(b)}$$
Which is well verified through numerical simulations.



Maybe I'm missing something obvious here, but I have absolutely no idea as of how to prove it. Uniform convergence is of course of no help here, the series $sumlimits_{n=0}^{infty}1$ being trivially divergent.



Any insight ?






real-analysis calculus sequences-and-series limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    A stronger result is also true. $sum_{n=0}^{infty} x^{b^n}+frac{ln(1-x)}{ln b}$ is boundedly oscillating as $xrightarrow 1-$.
    $endgroup$
    – i707107
    Jan 29 at 7:49


















8












$begingroup$


I conjecture that :
$$forall binmathbb{N}setminuslbrace0,1rbrace,limlimits_{xto 1^{-}}frac{1}{ln(1-x)}sumlimits_{n=0}^{infty}x^{b^n}=-frac{1}{ln(b)}$$
Which is well verified through numerical simulations.



Maybe I'm missing something obvious here, but I have absolutely no idea as of how to prove it. Uniform convergence is of course of no help here, the series $sumlimits_{n=0}^{infty}1$ being trivially divergent.



Any insight ?






real-analysis calculus sequences-and-series limits







share|cite|improve this question











$endgroup$












  • $begingroup$
    A stronger result is also true. $sum_{n=0}^{infty} x^{b^n}+frac{ln(1-x)}{ln b}$ is boundedly oscillating as $xrightarrow 1-$.
    $endgroup$
    – i707107
    Jan 29 at 7:49
















8












8








8


4



$begingroup$


I conjecture that :
$$forall binmathbb{N}setminuslbrace0,1rbrace,limlimits_{xto 1^{-}}frac{1}{ln(1-x)}sumlimits_{n=0}^{infty}x^{b^n}=-frac{1}{ln(b)}$$
Which is well verified through numerical simulations.



Maybe I'm missing something obvious here, but I have absolutely no idea as of how to prove it. Uniform convergence is of course of no help here, the series $sumlimits_{n=0}^{infty}1$ being trivially divergent.



Any insight ?






real-analysis calculus sequences-and-series limits







share|cite|improve this question











$endgroup$




I conjecture that :
$$forall binmathbb{N}setminuslbrace0,1rbrace,limlimits_{xto 1^{-}}frac{1}{ln(1-x)}sumlimits_{n=0}^{infty}x^{b^n}=-frac{1}{ln(b)}$$
Which is well verified through numerical simulations.



Maybe I'm missing something obvious here, but I have absolutely no idea as of how to prove it. Uniform convergence is of course of no help here, the series $sumlimits_{n=0}^{infty}1$ being trivially divergent.



Any insight ?






real-analysis calculus sequences-and-series limits




real-analysis calculus sequences-and-series limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 29 at 3:17







Harmonic Sun

















asked Jan 29 at 3:03









Harmonic SunHarmonic Sun

71210




71210












  • $begingroup$
    A stronger result is also true. $sum_{n=0}^{infty} x^{b^n}+frac{ln(1-x)}{ln b}$ is boundedly oscillating as $xrightarrow 1-$.
    $endgroup$
    – i707107
    Jan 29 at 7:49




















  • $begingroup$
    A stronger result is also true. $sum_{n=0}^{infty} x^{b^n}+frac{ln(1-x)}{ln b}$ is boundedly oscillating as $xrightarrow 1-$.
    $endgroup$
    – i707107
    Jan 29 at 7:49


















$begingroup$
A stronger result is also true. $sum_{n=0}^{infty} x^{b^n}+frac{ln(1-x)}{ln b}$ is boundedly oscillating as $xrightarrow 1-$.
$endgroup$
– i707107
Jan 29 at 7:49






$begingroup$
A stronger result is also true. $sum_{n=0}^{infty} x^{b^n}+frac{ln(1-x)}{ln b}$ is boundedly oscillating as $xrightarrow 1-$.
$endgroup$
– i707107
Jan 29 at 7:49












1 Answer
1






active

oldest

votes


















6












$begingroup$

Let $x = e^{-lambda}$, $lambda to 0^+$. We find that
$$
sum_{nge 0}x^{b^n} =sum_{nge 0}e^{-b^n lambda}=int_0^infty e^{-b^t lambda} mathrm{d}t + varepsilon_lambda,
$$ where $|varepsilon_lambda |le 1$ for all $lambda>0$, i.e. $varepsilon_lambda =O(1)$. By making substitution $b^tlambda =u$,
$$begin{eqnarray}
sum_{nge 0}e^{-b^n lambda}&=&frac{1}{ln b}int_lambda^infty e^{-u}frac{mathrm{d}u}{u}+O(1)\&=&frac{1}{ln b}int_lambda^1 frac{mathrm{d}u}{u}+frac{1}{ln b}int_lambda^infty frac{e^{-u}-1_{{ule 1}}}{u}mathrm{d}u+O(1)\
&=&-frac{ln lambda}{ln b}+O(1),
end{eqnarray}$$ since $$left|int_lambda^infty frac{e^{-u}-1_{{ule 1}}}{u}mathrm{d}uright|leint_0^infty frac{|e^{-u}-1_{{ule 1}}|}{u}mathrm{d}u<infty.$$ Finally, we have for all $b>1$,
$$begin{eqnarray}
lim_{xto 1^-} frac{1}{ln(1-x)}sumlimits_{nge 0}x^{b^n}&=&-frac{1}{ln b}lim_{lambda to 0^+}frac{ln lambda+O(1)}{ln(1-e^{-lambda})}\&=&-frac{1}{ln b}lim_{lambda to 0^+}frac{ln lambda}{ln(1-e^{-lambda})}\&=&-frac{1}{ln b}lim_{lambda to 0^+}frac{1/lambda}{e^{-lambda}/(1-e^{-lambda})}\&=&-frac{1}{ln b}lim_{lambda to 0^+}frac{e^lambda(1-e^{-lambda})}{lambda}=-frac{1}{ln b}.
end{eqnarray}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you, great answer !
    $endgroup$
    – Harmonic Sun
    Jan 29 at 12:00










  • $begingroup$
    Since you're the one who got the answer, you might be interested by my newest question here, which tries to refine this result even more.
    $endgroup$
    – Harmonic Sun
    Jan 29 at 13:30










  • $begingroup$
    @HarmonicSun Thank you for information :) I've left my answer to your new question !
    $endgroup$
    – Song
    Jan 29 at 13:43














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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









6












$begingroup$

Let $x = e^{-lambda}$, $lambda to 0^+$. We find that
$$
sum_{nge 0}x^{b^n} =sum_{nge 0}e^{-b^n lambda}=int_0^infty e^{-b^t lambda} mathrm{d}t + varepsilon_lambda,
$$ where $|varepsilon_lambda |le 1$ for all $lambda>0$, i.e. $varepsilon_lambda =O(1)$. By making substitution $b^tlambda =u$,
$$begin{eqnarray}
sum_{nge 0}e^{-b^n lambda}&=&frac{1}{ln b}int_lambda^infty e^{-u}frac{mathrm{d}u}{u}+O(1)\&=&frac{1}{ln b}int_lambda^1 frac{mathrm{d}u}{u}+frac{1}{ln b}int_lambda^infty frac{e^{-u}-1_{{ule 1}}}{u}mathrm{d}u+O(1)\
&=&-frac{ln lambda}{ln b}+O(1),
end{eqnarray}$$ since $$left|int_lambda^infty frac{e^{-u}-1_{{ule 1}}}{u}mathrm{d}uright|leint_0^infty frac{|e^{-u}-1_{{ule 1}}|}{u}mathrm{d}u<infty.$$ Finally, we have for all $b>1$,
$$begin{eqnarray}
lim_{xto 1^-} frac{1}{ln(1-x)}sumlimits_{nge 0}x^{b^n}&=&-frac{1}{ln b}lim_{lambda to 0^+}frac{ln lambda+O(1)}{ln(1-e^{-lambda})}\&=&-frac{1}{ln b}lim_{lambda to 0^+}frac{ln lambda}{ln(1-e^{-lambda})}\&=&-frac{1}{ln b}lim_{lambda to 0^+}frac{1/lambda}{e^{-lambda}/(1-e^{-lambda})}\&=&-frac{1}{ln b}lim_{lambda to 0^+}frac{e^lambda(1-e^{-lambda})}{lambda}=-frac{1}{ln b}.
end{eqnarray}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you, great answer !
    $endgroup$
    – Harmonic Sun
    Jan 29 at 12:00










  • $begingroup$
    Since you're the one who got the answer, you might be interested by my newest question here, which tries to refine this result even more.
    $endgroup$
    – Harmonic Sun
    Jan 29 at 13:30










  • $begingroup$
    @HarmonicSun Thank you for information :) I've left my answer to your new question !
    $endgroup$
    – Song
    Jan 29 at 13:43


















6












$begingroup$

Let $x = e^{-lambda}$, $lambda to 0^+$. We find that
$$
sum_{nge 0}x^{b^n} =sum_{nge 0}e^{-b^n lambda}=int_0^infty e^{-b^t lambda} mathrm{d}t + varepsilon_lambda,
$$ where $|varepsilon_lambda |le 1$ for all $lambda>0$, i.e. $varepsilon_lambda =O(1)$. By making substitution $b^tlambda =u$,
$$begin{eqnarray}
sum_{nge 0}e^{-b^n lambda}&=&frac{1}{ln b}int_lambda^infty e^{-u}frac{mathrm{d}u}{u}+O(1)\&=&frac{1}{ln b}int_lambda^1 frac{mathrm{d}u}{u}+frac{1}{ln b}int_lambda^infty frac{e^{-u}-1_{{ule 1}}}{u}mathrm{d}u+O(1)\
&=&-frac{ln lambda}{ln b}+O(1),
end{eqnarray}$$ since $$left|int_lambda^infty frac{e^{-u}-1_{{ule 1}}}{u}mathrm{d}uright|leint_0^infty frac{|e^{-u}-1_{{ule 1}}|}{u}mathrm{d}u<infty.$$ Finally, we have for all $b>1$,
$$begin{eqnarray}
lim_{xto 1^-} frac{1}{ln(1-x)}sumlimits_{nge 0}x^{b^n}&=&-frac{1}{ln b}lim_{lambda to 0^+}frac{ln lambda+O(1)}{ln(1-e^{-lambda})}\&=&-frac{1}{ln b}lim_{lambda to 0^+}frac{ln lambda}{ln(1-e^{-lambda})}\&=&-frac{1}{ln b}lim_{lambda to 0^+}frac{1/lambda}{e^{-lambda}/(1-e^{-lambda})}\&=&-frac{1}{ln b}lim_{lambda to 0^+}frac{e^lambda(1-e^{-lambda})}{lambda}=-frac{1}{ln b}.
end{eqnarray}$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you, great answer !
    $endgroup$
    – Harmonic Sun
    Jan 29 at 12:00










  • $begingroup$
    Since you're the one who got the answer, you might be interested by my newest question here, which tries to refine this result even more.
    $endgroup$
    – Harmonic Sun
    Jan 29 at 13:30










  • $begingroup$
    @HarmonicSun Thank you for information :) I've left my answer to your new question !
    $endgroup$
    – Song
    Jan 29 at 13:43
















6












6








6





$begingroup$

Let $x = e^{-lambda}$, $lambda to 0^+$. We find that
$$
sum_{nge 0}x^{b^n} =sum_{nge 0}e^{-b^n lambda}=int_0^infty e^{-b^t lambda} mathrm{d}t + varepsilon_lambda,
$$ where $|varepsilon_lambda |le 1$ for all $lambda>0$, i.e. $varepsilon_lambda =O(1)$. By making substitution $b^tlambda =u$,
$$begin{eqnarray}
sum_{nge 0}e^{-b^n lambda}&=&frac{1}{ln b}int_lambda^infty e^{-u}frac{mathrm{d}u}{u}+O(1)\&=&frac{1}{ln b}int_lambda^1 frac{mathrm{d}u}{u}+frac{1}{ln b}int_lambda^infty frac{e^{-u}-1_{{ule 1}}}{u}mathrm{d}u+O(1)\
&=&-frac{ln lambda}{ln b}+O(1),
end{eqnarray}$$ since $$left|int_lambda^infty frac{e^{-u}-1_{{ule 1}}}{u}mathrm{d}uright|leint_0^infty frac{|e^{-u}-1_{{ule 1}}|}{u}mathrm{d}u<infty.$$ Finally, we have for all $b>1$,
$$begin{eqnarray}
lim_{xto 1^-} frac{1}{ln(1-x)}sumlimits_{nge 0}x^{b^n}&=&-frac{1}{ln b}lim_{lambda to 0^+}frac{ln lambda+O(1)}{ln(1-e^{-lambda})}\&=&-frac{1}{ln b}lim_{lambda to 0^+}frac{ln lambda}{ln(1-e^{-lambda})}\&=&-frac{1}{ln b}lim_{lambda to 0^+}frac{1/lambda}{e^{-lambda}/(1-e^{-lambda})}\&=&-frac{1}{ln b}lim_{lambda to 0^+}frac{e^lambda(1-e^{-lambda})}{lambda}=-frac{1}{ln b}.
end{eqnarray}$$






share|cite|improve this answer











$endgroup$



Let $x = e^{-lambda}$, $lambda to 0^+$. We find that
$$
sum_{nge 0}x^{b^n} =sum_{nge 0}e^{-b^n lambda}=int_0^infty e^{-b^t lambda} mathrm{d}t + varepsilon_lambda,
$$ where $|varepsilon_lambda |le 1$ for all $lambda>0$, i.e. $varepsilon_lambda =O(1)$. By making substitution $b^tlambda =u$,
$$begin{eqnarray}
sum_{nge 0}e^{-b^n lambda}&=&frac{1}{ln b}int_lambda^infty e^{-u}frac{mathrm{d}u}{u}+O(1)\&=&frac{1}{ln b}int_lambda^1 frac{mathrm{d}u}{u}+frac{1}{ln b}int_lambda^infty frac{e^{-u}-1_{{ule 1}}}{u}mathrm{d}u+O(1)\
&=&-frac{ln lambda}{ln b}+O(1),
end{eqnarray}$$ since $$left|int_lambda^infty frac{e^{-u}-1_{{ule 1}}}{u}mathrm{d}uright|leint_0^infty frac{|e^{-u}-1_{{ule 1}}|}{u}mathrm{d}u<infty.$$ Finally, we have for all $b>1$,
$$begin{eqnarray}
lim_{xto 1^-} frac{1}{ln(1-x)}sumlimits_{nge 0}x^{b^n}&=&-frac{1}{ln b}lim_{lambda to 0^+}frac{ln lambda+O(1)}{ln(1-e^{-lambda})}\&=&-frac{1}{ln b}lim_{lambda to 0^+}frac{ln lambda}{ln(1-e^{-lambda})}\&=&-frac{1}{ln b}lim_{lambda to 0^+}frac{1/lambda}{e^{-lambda}/(1-e^{-lambda})}\&=&-frac{1}{ln b}lim_{lambda to 0^+}frac{e^lambda(1-e^{-lambda})}{lambda}=-frac{1}{ln b}.
end{eqnarray}$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 29 at 5:22

























answered Jan 29 at 5:16









SongSong

18.5k21651




18.5k21651












  • $begingroup$
    Thank you, great answer !
    $endgroup$
    – Harmonic Sun
    Jan 29 at 12:00










  • $begingroup$
    Since you're the one who got the answer, you might be interested by my newest question here, which tries to refine this result even more.
    $endgroup$
    – Harmonic Sun
    Jan 29 at 13:30










  • $begingroup$
    @HarmonicSun Thank you for information :) I've left my answer to your new question !
    $endgroup$
    – Song
    Jan 29 at 13:43




















  • $begingroup$
    Thank you, great answer !
    $endgroup$
    – Harmonic Sun
    Jan 29 at 12:00










  • $begingroup$
    Since you're the one who got the answer, you might be interested by my newest question here, which tries to refine this result even more.
    $endgroup$
    – Harmonic Sun
    Jan 29 at 13:30










  • $begingroup$
    @HarmonicSun Thank you for information :) I've left my answer to your new question !
    $endgroup$
    – Song
    Jan 29 at 13:43


















$begingroup$
Thank you, great answer !
$endgroup$
– Harmonic Sun
Jan 29 at 12:00




$begingroup$
Thank you, great answer !
$endgroup$
– Harmonic Sun
Jan 29 at 12:00












$begingroup$
Since you're the one who got the answer, you might be interested by my newest question here, which tries to refine this result even more.
$endgroup$
– Harmonic Sun
Jan 29 at 13:30




$begingroup$
Since you're the one who got the answer, you might be interested by my newest question here, which tries to refine this result even more.
$endgroup$
– Harmonic Sun
Jan 29 at 13:30












$begingroup$
@HarmonicSun Thank you for information :) I've left my answer to your new question !
$endgroup$
– Song
Jan 29 at 13:43






$begingroup$
@HarmonicSun Thank you for information :) I've left my answer to your new question !
$endgroup$
– Song
Jan 29 at 13:43




















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