1
$\begingroup$

Let $-1<t<0$. Is there $t$ such that for all positive integers $n>c$($c$ is a positive integer depends on $t$)

$n^t - (n+1)^t > 1/n$

$\endgroup$
2
  • $\begingroup$ It is possible to say so if it's true. Is it ? t has to be negative. The t you picked is clearly not $\endgroup$
    – T_O
    Mar 27, 2014 at 9:39
  • $\begingroup$ I deleted it. It was just an example. The point is, is there a such $t$ ? $\endgroup$
    – esege
    Mar 27, 2014 at 9:41

2 Answers 2

2
$\begingroup$

Yes, there is. Let $t=-s$ with $0<s<1$. Note that for all $n>1$ we have $n>n^s\iff \frac1{n^s}>\frac1n\iff n^t>\frac1n$. It follows that $n^t+(n+1)^t>n^t>\frac1n$

$\endgroup$
2
$\begingroup$

Yes, for example if you take $t=-\frac{1}{2}$, then your inequality holds for all n>.5 (in fact its something like .419643). I used wolfram alpha for the computation, you can check it out yourself.

http://www.wolframalpha.com/input/?i=1%2Fsqrt%28n%29%2B1%2Fsqrt%28n%2B1%29%3E1%2Fn

$\endgroup$
1
  • $\begingroup$ Thank you. But it was my mistake. It's not $n^t + (n+1)^t > 1/n$ . It's $n^t - (n+1)^t > 1/n$ $\endgroup$
    – esege
    Mar 27, 2014 at 9:53

You must log in to answer this question.

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