1
$\begingroup$

I'm trying to prove the following inequality. Let $ s = x+iy \in \mathbb{C} $. Prove that

$$ \left\lvert\frac{1}{n^s} - \frac{1}{(n+1)^s}\right\rvert \leq \frac{\lvert s\rvert}{n^{x+1}} $$

The problem is I'm not very familiar with complex analysis. I've tried to write it down with the definition $ n^s = e^{s\log{n}} = e^{x\log{n}}(\cos(y\log{n}) + i\sin(y\log{n})) $, but it has lead me nowhere

$\endgroup$
2
  • $\begingroup$ $e^{iy} = \cos(y)+\color{red}{i}\sin(y)$ $\endgroup$
    – hjpotter92
    Nov 30, 2014 at 14:17
  • $\begingroup$ Of course, thanks $\endgroup$
    – Jytug
    Nov 30, 2014 at 14:33

1 Answer 1

1
$\begingroup$

The fundamental theorem of calculus tells you that

$$f(n+1) - f(n) = \int_n^{n+1} f'(t)\,dt$$

for continuously differentiable functions $f\colon (0,+\infty) \to \mathbb{C}$. You get the estimate

$$\lvert f(n+1) - f(n)\rvert \leqslant \sup \{ \lvert f'(t)\rvert : t \in [n,n+1]\}.$$

All that you then need is to find

$$\lvert t^z\rvert$$

for real positive $t$ and complex $z$.

$\endgroup$
1
  • $\begingroup$ Thanks a lot, I didn't expect to use that theorem $\endgroup$
    – Jytug
    Nov 30, 2014 at 17:18

You must log in to answer this question.

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