A **proof** for $\sum_{i=0}^{t-2}{\frac{1}{t+3i}} \leq \frac{1}{2}$

Question

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 $t$?

Answer 1

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 $0.5$

Answer 2

Might be a bit overkill but well...

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