Question. Numerically, the following is convincing. However, is there a proof? $$\left(\sum_{k\geq1}\frac1{\sqrt{2^k+3^k}}\right)^4 <\pi^2\left(\sum_{k\geq1}\frac1{2^k+3^k}\right)\left(\sum_{k\geq1}\frac{k^2}{2^k+3^k}\right).$$

This comes up in some recent work and the inequality seems needed.

  • 5
    $\begingroup$ It seems you only need 3 terms for each sum on the right and a crude upper bound for the sum on the left. $\endgroup$ – Brendan McKay Feb 12 '17 at 7:27
  • 3
    $\begingroup$ This is a research-level question? $\endgroup$ – user541686 Feb 13 '17 at 0:22

This may serve as a different approach. By Cauchy-Schwarz inequality, $$\left(\sum_{k\geq 1}\frac 1{\sqrt{2^k+3^k}}\right)^2\leq \left(\sum_{k\geq 1}\frac 1{k^2}\right)\left(\sum_{k\geq 1}\frac{k^2}{2^k+3^k}\right),$$ which shows that $$\left(\sum_{k\geq 1}\frac 1{\sqrt{2^k+3^k}}\right)^2\leq \frac{\pi^2}6\left(\sum_{k\geq 1}\frac{k^2}{2^k+3^k}\right).$$

So it suffices to show that $$\left(\sum_{k\geq 1}\frac 1{\sqrt{2^k+3^k}}\right)^2<6\cdot \sum_{k\geq 1}\frac 1 {2^k+3^k}.$$

But \begin{align} \left(\sum_{k\geq 1}\frac 1{\sqrt{2^k+3^k}}\right)^2 &<\left(\sum_{k\geq 1}\frac 1{\sqrt{2\cdot 6^{k/2}}}\right)^2 \\ &<6\left(\frac 1{2+3}+\sum_{k\geq 2}\frac 1{2\cdot 3^k}\right) \\ &<6\left(\sum_{k\geq 1}\frac 1{2^k+3^k}\right), \end{align} where in the first inequality, the AM-GM inequality was used.

  • 1
    $\begingroup$ Second "<" in the penultimate line of your proof, hmm, I don't catch it: Could you hint at how to get to factor "6" et al? $\endgroup$ – Hanno Feb 15 '17 at 7:44
  • 1
    $\begingroup$ @Hanno and Cherng-tiao Perng: The second inequality in the last derivation is wrong. Both sides are geometric series the values of which violate the inequality. $\endgroup$ – Hans Jan 31 '18 at 23:20

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.