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).


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.

  • $\begingroup$ @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

Begin with the following, obtained from wolfram:

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


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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.