Is this inequality true ?

  • #1
Is this inequality true ??

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

Is it true ??

Thanks
 

Answers and Replies

  • #2
disregardthat
Science Advisor
1,861
34


Can you write [tex]\frac{1}{(k+1)^2(n-k+1)}[/tex] differently? You want something on the form [tex]\frac{a}{k+1} + \frac{bk+c}{(k+1)^2} + \frac{d}{n-k+1}[/tex] .
 
  • #3


It really work thanks .
 
Last edited:

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