0
$\begingroup$

The problem is from my approach to $\sum\limits_{n=1}^\infty (-1)^{n-1}\left(\alpha-\frac{(n-\beta)^n}{n^n}\right)$. There's pretty much it. It can be assumed $n\in\mathbb{N}$ if needed.
My effort: done some numerical research that shows the inequality holds for $n>a>0$ and for $a<0<n$. Still not a clue how to tackle the inequality.

$\endgroup$

1 Answer 1

2
$\begingroup$

Using the Bernoulli inequality is appropriate as $\forall a \in \mathbb{R} \exists N \in \mathbb{N}: n > N \Rightarrow \frac{a}{(n-a) (n+1)} > -1$

I.e. $\left(1+\frac{a}{(n-a) (n+1)}\right)^n(1-\frac{a}{n+1}) \geq \left(1+\frac{n}{(n-a)}\frac{a}{ (n+1)}\right)(1-\frac{a}{n+1}) = 1 + \frac{a}{n+1}\frac{n}{n-a} - \frac{a}{n+1}- \big(\frac{a}{n+1}\big)^2\frac{n}{n-a} = 1 + \frac{a}{n+1}\frac{n-(n-a)}{n-a} - \big(\frac{a}{n+1}\big)^2\frac{n}{n-a} = 1 + \frac{a}{n+1}\frac{a}{n-a} - \big(\frac{a}{n+1}\big)^2\frac{n}{n-a} = 1 + \frac{a^2}{(n+1)(n-a)}\big(1-\frac{n}{n+1}\big) > 1$ for sufficiently large $n$, as in such cases $\frac{a^2}{(n+1)(n-a)} > 0$ and $\big(1-\frac{n}{n+1}\big)$ converges to zero from above.

$\endgroup$
1
  • $\begingroup$ Yep, Bernoulli inequality, that's it, thanks. $=1+\frac{1}{(n + 1)^2 (n - a)}$ would be sufficent. $\endgroup$ May 21, 2020 at 2:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .