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I need a proof for the inequality: $\sum_{i=0}^{t-2}{\frac{1}{t+3i}} \leq \frac{1}{2}$ for all natural numbers $t \geq 2$. For $t=2$ both sides are equal.

Can someone find a proof for all $t$? maybe by proving monotonicity (non-increasing) in $t$?

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Might be a bit overkill but well...

Check it is true for $t < 27$ or so. Then assuming $t \geq 27\ (*)$ : $$ \sum_{i=0}^{t-2} \frac{1}{t+3i} = \frac{1}{t} + \sum_{i=1}^{t-2} \frac{1} {t+3i} \leq \frac{1}{t} + \int_0^{t-2} \frac{1}{t+3x} \ dx = \frac{1}{t} + \frac{1}{3} log(4 - 6/t) \leq \frac{1}{t} + \frac{1}{3}log(4) \overset{(*)}{\leq} \frac{1}{2} $$

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For $t=2.25$ the sum seems to be greater than $\dfrac{1}{2}$ The sum has a maximum between $0.368\lt t\lt 0.369$ and I think is greater than $0.5$

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  • $\begingroup$ I think $t$ is a natural number. $\endgroup$ – Dragonemperor42 Jul 12 '16 at 10:30
  • $\begingroup$ @Roby5: in this case it needs to specify $t\in\mathbb{N}$ in the question $\endgroup$ – Riccardo.Alestra Jul 12 '16 at 10:39
  • $\begingroup$ How do you sum from 0 to $t-2$ when $t$ is not natural? $\endgroup$ – Ophir Friedler Jul 12 '16 at 10:41
  • $\begingroup$ @user1247378: Ok. It could solve the problem $\endgroup$ – Riccardo.Alestra Jul 12 '16 at 10:44
  • $\begingroup$ @Riccardo.Alestra Do you have a suggestion how? :) Proof-wise $\endgroup$ – Ophir Friedler Jul 12 '16 at 10:44

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