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consider the folloing sum

$$ \sum_{n=0}^\infty a_ne^{-\alpha_nx} $$

for every n $ 0 < \alpha_n < \alpha_{n+1}$

it is also given that for $x_0 \in \Bbb{R}$ the sum converges

prove that the sum converges uniformly in the interval $[x_0,\infty]$

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  • $\begingroup$ Weierstrass and the fact that $e^{-\alpha_n x}\leq e^{-\alpha_n x_0}$ $\endgroup$ – Antoine Aug 11 '15 at 18:49
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    $\begingroup$ @Antoine Only if $a_n\ge 0$ $\endgroup$ – David C. Ullrich Aug 11 '15 at 19:57
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    $\begingroup$ You can do this using summation by parts. en.wikipedia.org/wiki/Summation_by_parts $\endgroup$ – David C. Ullrich Aug 11 '15 at 20:09
  • $\begingroup$ @hermes Why is $a_n \geq 0$ presumed? (I missed that and this may be the reason that you have a down vote) $\endgroup$ – Antoine Aug 12 '15 at 14:05
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    $\begingroup$ Best not to use $\alpha_n$ and $a_n$ in the same question, when you can avoid it. $\endgroup$ – Robert Israel Aug 13 '15 at 0:44
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First we prove it is true for $x_0=0$. The given condition becomes $\sum_{k=0}^n a_k$ converges. The idea is to use partial summation (used in proving Abel theorem) to get an estimate of Cauchy sum.

Let $b_n=\sum_{k=m}^n a_k$. So by Cauchy Criterion, $|b_n|<\epsilon$ for $n,m>N$.

We have \begin{align} \sum_{k=m}^n a_ke^{-\alpha_kx}&=\sum_{k=m}^n (b_k-b_{k-1})e^{-\alpha_kx} \\ &=\sum_{k=m}^n b_ke^{-\alpha_kx} -\sum_{k=m}^n b_{k-1}e^{-\alpha_kx} \\ &=\sum_{k=m}^{n-1} b_k(e^{-\alpha_kx}-e^{-\alpha_{k+1}x})+b_ne^{-\alpha_nx}\tag{$b_{m−1}=0$} \end{align} Since $\alpha_n>0, \:\alpha_n \uparrow$, $e^{-\alpha_kx}-e^{-\alpha_{k+1}x}\geqslant0\:$ for all $k>0$ and $x\in[0,\infty)$.

Since $-\epsilon<b_k<\epsilon$ for all $k>m$ $$ |b_k(e^{-\alpha_kx}-e^{-\alpha_{k+1}x})|<\epsilon(e^{-\alpha_kx}-e^{-\alpha_{k+1}x}) $$ So for all $n,m>N-1$ and $x\in[0,\infty)$, there is \begin{align} \left|\sum_{k=m}^n a_ke^{-\alpha_kx}\right|&\leqslant\sum_{k=m}^{n-1} |b_k(e^{-\alpha_kx}-e^{-\alpha_{k+1}x})|+|b_ne^{-\alpha_nx}| \\ &\leqslant\sum_{k=m}^{n-1} \epsilon(e^{-\alpha_kx}-e^{-\alpha_{k+1}x})+\epsilon e^{-\alpha_nx} \\ &=\epsilon \:(e^{-\alpha_{m}x}-e^{-\alpha_{n}x}+e^{-\alpha_{n}x}) \\ &=\epsilon \:e^{-\alpha_{m}x} \\ &\leqslant\epsilon \end{align} So by Cauchy Criterion, it uniformly converges on $[0,\infty)$. Finally we can prove it on $[x_0,\infty)$ by replace $x$ with $x-x_0$.

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  • $\begingroup$ didnt get this part $\sum_{k=m}^n b_ke^{-\alpha_kx} -\sum_{k=m}^n b_{k-1}e^{-\alpha_kx} = \sum_{k=m}^{n-1} b_k(e^{-\alpha_kx}-e^{-\alpha_{k+1}x})+b_ne^{-\alpha_nx}$ arent you missing $ + b_{m-1}e^{-\alpha_mx} $? $\endgroup$ – Daniel Katzan Aug 13 '15 at 7:45
  • $\begingroup$ also didnt get this equality $\sum_{k=m}^{n-1} \epsilon(e^{-\alpha_kx}-e^{-\alpha_{k+1}x})+\epsilon e^{-\alpha_nx} =\epsilon \:e^{-\alpha_{m}x} $ $\endgroup$ – Daniel Katzan Aug 13 '15 at 7:46
  • $\begingroup$ By definition $b_{m−1}=0$. 2nd question left is telescope series. Read the proof again. $\endgroup$ – Math Wizard Aug 13 '15 at 8:49

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