Non negativity involving sequences

Question

Define for $n\in\mathbb{N}$ $$a_n=\left[n\sum_{k=1}^{n}\frac{1}{k^5}\right]$$ where $[x]$ denotes the greatest integer $\leq x$.

Prove that $$(n+1)a_n-n a_{n+1}+1\geq 0 \ \ \forall n\geq 1$$

$$(n+1)a_n-n a_{n+1}+1=(n+1)\left[n\sum_{k=1}^{n}\frac{1}{k^5}\right]-n\left[(n+1)\sum_{k=1}^{n+1}\frac{1}{k^5}\right]+1$$ Now using $[x]=x-\{x\}$ we have $$(n+1)a_n-n a_{n+1}+1=(n+1)\left(n\sum_{k=1}^{n}\frac{1}{k^5}\right)-(n+1)\left\{n\sum_{k=1}^{n}\frac{1}{k^5}\right\}-n\left((n+1)\sum_{k=1}^{n+1}\frac{1}{k^5}\right)+n\left\{(n+1)\sum_{k=1}^{n+1}\frac{1}{k^5}\right\}+1$$ So we get $$(n+1)a_n-n a_{n+1}+1=n\left\{(n+1)\sum_{k=1}^{n+1}\frac{1}{k^5}\right\}-(n+1)\left\{n\sum_{k=1}^{n}\frac{1}{k^5}\right\}-\frac{n}{(n+1)^4}+1$$ Now we have $$\sum_{k=1}^{n}\frac{1}{k^5}=H_n^{(5)}$$ where $H_n^{(5)}$ is the generalised harmonic number. So we obtain $$(n+1)a_n-n a_{n+1}+1=n\left\{(n+1)H_{n+1}^{(5)}\right\}-(n+1)\left\{nH_n^{(5)}\right\}-\frac{n}{(n+1)^4}+1$$ Now we have using Wolfram Alpha $$(n+1)a_n-n a_{n+1}+1\geq 0\ \ \forall 1\leq n \leq 50$$ Also we have using Wolfram Alpha $$(n+1)a_n-n a_{n+1}+1\geq 0\ \ \forall 50\leq n \leq 100$$ I need a proof for any $n\in\mathbb{N}$. Thank you for your time.

Answer 1

In short, it's false. The counterexamples lie near $$n = n_k = \left[\frac{k}{\zeta(5)-1}\right]$$ for any integer $k > 0$.

For instance, $k=1$ gives $n=27$, where your inequality becomes $-26 \geq 0$.

By the way, changing your wolfram alpha search slightly might have told you this already.