$\lim_{n\to\infty}\frac{\sum_{i=1}^ni^{4\lambda}}{\Big(\sum_{i=1}^ni^{2\lambda}\Big)^2}=0$

Question

Is it true that $\forall \lambda>0$ $$\lim_{n\to\infty}\frac{\sum_{i=1}^ni^{4\lambda}}{\Big(\sum_{i=1}^ni^{2\lambda}\Big)^2}=0$$ I cannot find a way to prove it, nor can I find a counterexample. Any help is greatly appreciated!

Answer 1

The numerator is less than $n\cdot n^{4\lambda}.$ In the denominator there are at least $n/2$ terms that are at least $(n/2)^{2\lambda}.$ Thus our expression is less than

$$\frac{n\cdot n^{4\lambda}}{[(n/2)(n/2)^{2\lambda}]^2}.$$

This is on the order of $\dfrac{1}{n}$ as $n\to \infty,$ hence the limit is $0.$