The following two sums I evaluated using mathematica $$ S_1 = \sum_{i = 0}^\infty \left(\frac{1}{4}\right)^i = \frac{4}{3} $$ and the sum $$ S_2 =\sum_{i = 0}^\infty \left(\frac{3}{4}\right)^i = 4 $$ This seems really strange to me as $S_2$ should be larger than $3S_1$. This is easily motivated by looking at the first couple of terms of $S_2$ \begin{align} S_2 &= 3\left(\frac{1}{4}\right) + 3\left(\frac{1}{4}\right)^2 3 + 3\left(\frac{1}{4}\right)^3 3^2 +\cdots \\ &= 3\left[\left(\frac{1}{4}\right) + \left(\frac{1}{4}\right)^2 3 + \left(\frac{1}{4}\right)^3 3^2 +\cdots\right] \\ &> 3 S_1 \end{align} Can somebody explain me either if Mathematica is wrong or how these are the correct expressions for the sums?
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You are ignoring the terms corresponding to $i=0$ and that makes a difference.
Kavi Rama Murthy
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You are correct that $$ \sum_{i=1}^\infty \left(\frac34\right)^i>3\sum_{i=1}^\infty\left(\frac14\right)^i.$$ However, note that the LHS is $3$ and RHS is $1$, instead of $4$ and $4$ as you claimed. Your mistake was missing out the $i=0$ term, as the other answer also explains.
Once you include this term, your "proof" that $S_2>3S_1$ doesn't work anymore, because $3\times 1>1$.
YiFan Tey
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