Flipping Everything Upside Down

Algebra Level 2

$\large A = \sum_{k=0}^\infty \dfrac1{\sqrt[3]{2^k}} \qquad \qquad B = \sum_{k=0}^\infty \dfrac1{\sqrt{2^k}}$

What is the relationship between $A$ and $B$ ?

It is impossible to determine

A>B

A<B

A=B

4 solutions

Sam Bealing
Jul 1, 2016

As $k \geq 0$ it follows that $2^k \geq 1 \implies \sqrt[3]{2^k} \leq \sqrt{2^k} \implies \dfrac{1}{ \sqrt[3]{2^k}} \geq \dfrac{1}{\sqrt{2^k}}$ .

For $k>0$ we have $2^k >1$ and hence the inequality becomes strict so we have each term of $A$ is greater than the same term in $B$ .

$\boxed{\boxed{A>B}}$

Moderator note:

Good approach comparing termwise.

M K
Jun 30, 2016

Hm, there are several issues with your first equation

Numerator is missing a $a$
Denominator is just $1 - r$

Calvin Lin Staff - 4 years, 11 months ago

Whoops, my bad. Thanks for the correction!

M K - 4 years, 11 months ago

Typo: $\dfrac{1}{\sqrt[3]{2}} > \dfrac{1}{\sqrt{2}}$

Hung Woei Neoh - 4 years, 11 months ago

Joe Previdi
Jul 3, 2016

Since since every term which appears in B also appears in A (every time k is a multiple of 3) and there are more terms than just B's terms, A has every term of B plus more, and therefore A is greater.

Jeannine Myer
Jul 3, 2016

If you write out the sum, you see between the terms which are equal you add a bigger number each time with A. Simple.

Flipping Everything Upside Down

4 solutions

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