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Is it true that $\displaystyle \int_0^{\infty} 2^{-x}\,dx=\sum_{x=0}^{\infty}\lim_{k\to\infty}\dfrac{2^{-x}}{k}$?

Here is my try:

\begin{align} & \sum_{x=0}^{\infty}\lim_{k\to\infty}\dfrac{2^{-x}}{k}\\ &= \lim_{k\to\infty}\sum_{x=0}^{\infty}\dfrac{2^{-x}}{k} \\ &= \lim_{h\to0+}\sum_{x=0}^{\infty}2^{-x}h \\ &= \int_0^{\infty} 2^{-x}\,dx \\ &= \frac{1}{\log{2}} \end{align}

Is it correct? I have doubts because doing

\begin{align} & \sum_{x=0}^{\infty}\lim_{k\to\infty}\dfrac{2^{-x}}{k}\\ &= \lim_{k\to\infty}\dfrac 1k\sum_{x=0}^{\infty}2^{-x} \\ &= \lim_{k\to\infty}\dfrac 2k \\ &= 0 \neq \frac 1{\log 2} \end{align}

I get another answer from the same equation.

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    $\begingroup$ no,RHS is equal to zero $\endgroup$ – haqnatural Nov 28 '17 at 19:50
  • $\begingroup$ What is $\sum_{x=0}^\infty $ $\endgroup$ – hamam_Abdallah Nov 28 '17 at 19:56
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The fallacy is in assuming $\int_0^\infty f(x) dx=\lim_{h\to 0^+}\sum_{n=0}^\infty hf(n)$; the right-hand side should be $\lim_{h\to 0^+}\sum_{n=0}^\infty \frac{h}{2}(f(hn)+f(hn+h))$. You can convince yourself of this by drawing thin trapezia under a curve.

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