-2
$\begingroup$

Given $N$ and $a$ positive integers, with $a\ge 2$ is it possible to prove the inequality: $$\sum_{k=1}^N\frac{k^a}{(k+1)^a+(k+2)^a}\le\frac{N}{2}$$

$\endgroup$
2
  • 2
    $\begingroup$ I don't know if this one is appropiate on this site. As $\frac{1}{N} \cdot \sum_{k=1}^N \dots $ converges to $\frac{1}{2}$ you could show that it is monotone increasing and you are done $\endgroup$ Commented May 21, 2013 at 13:11
  • $\begingroup$ You can use the principle of mathematical induction to prove such inequalities. $\endgroup$
    – aslan
    Commented May 21, 2013 at 14:11

1 Answer 1

2
$\begingroup$

Since $\displaystyle \frac{k^a}{(k+1)^a+(k+2)^a}<\frac{k^a}{k^a+k^a}=\frac{1}{2}$ then $\displaystyle\sum_{k=1}^N\frac{k^a}{(k+1)^a+(k+2)^a}<\sum_{k=1}^N\frac{1}{2}=\frac{N}{2}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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