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I saw the following inequality as part of a proof of a theorem and I can not figure out how one go from the left hand side to the right.

$$\sum_{j=0}^k \sqrt{2^{j+1}}<\sqrt{2^{k+1}}\sum_{i=0}^{\infty}2^{-\frac{i}{2}}$$

Edit: removed the unnecessary 2 and ln(n).

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It's $$\sum_{j=0}^k(\sqrt2)^{j+1}<(\sqrt2)^{k+1}\cdot\frac{1}{1-\frac{1}{\sqrt2}}$$ or $$\frac{\sqrt2((\sqrt2)^{k+1}-1)}{\sqrt2-1}<(\sqrt2)^{k+1}\cdot\frac{\sqrt2}{\sqrt2-1}$$ or $$(\sqrt2)^{k+1}-1<(\sqrt2)^{k+1},$$ which is obvious.

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  • $\begingroup$ @ChuckP The left side it's $S_n=\frac{a_1(q^n-1)}{q-1}$ for geometric progression. Here $a_1=\sqrt2$, $n=k+1$ and $q=\sqrt2$. The right side it's $S=\frac{a_1}{1-q}$, where $a_1=1$ and $q=\frac{1}{\sqrt2}$. $\endgroup$ – Michael Rozenberg Oct 22 '17 at 14:56
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Begin with the following, obtained from wolfram:

$\sum_{j=0}^{k}\sqrt{2^{j+1}} = (1+\sqrt{2})(2\sqrt{2^{k}} - \sqrt{2})$

and

$\sum_{i=0}^{\infty}2^{-\frac{i}{2}} = (2+\sqrt{2})$

We now have: $$(1+\sqrt{2})(2\sqrt{2^{k}} - \sqrt{2}) \lt \sqrt{2^{k+1}}(2+\sqrt{2})$$


Re-express each side by means of exponential manipulation:

L.H.S. $(1+\sqrt{2})(2\sqrt{2^{k}} - \sqrt{2}) = (1+\sqrt{2})(\sqrt{2^{k+2}} - \sqrt{2})$

R.H.S. $\sqrt{2^{k+1}}(2+\sqrt{2}) = \sqrt{2^{k+2}}(\sqrt{2} + 1)$


Removing $(1+\sqrt{2})$ from both sides yields

$$\sqrt{2^{k+2}} - \sqrt{2} \lt \sqrt{2^{k+2}}$$

which is obviously true.

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