You're welcome! Glad it worked for you.

[mahmoud2011]

Is this inequality true ??

$0\leq \sum_{k=0}^{n} \frac{1}{(k+1)^2 (n-k+1)} \leq \frac{1}{\sqrt{n+1}}$ for all natural numbers n

Is it true ??

Thanks

 

[disregardthat]

Can you write $$\frac{1}{(k+1)^2(n-k+1)}$$ differently? You want something on the form $$\frac{a}{k+1} + \frac{bk+c}{(k+1)^2} + \frac{d}{n-k+1}$$ .

 

[mahmoud2011]

It really work thanks .