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I'm struggling with proving the following inequality: $$\sum_{i=0}^{t-2}{\frac{1}{t+3i}} \leq \frac{1}{2}$$

for all $t \geq 2$. I think it is monotonic non-increasing in $t$, which would suffice. Thanks in advance!

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Let $N=t-2$. We have: $$ \sum_{i=0}^{t-2}\frac{1}{t+3i}=S_N=\sum_{k=0}^{N}\frac{1}{N+3k+2}=\frac{1}{N+2}\sum_{k=0}^{N}\frac{1}{1+3\frac{k}{N+2}} $$ and $S_N$ is greater than $$ \int_{0}^{1}\frac{dx}{1+3x} = \frac{2}{3}\log(2) =0.46209812\ldots$$ by the Hermite-Hadamard inequality, but $S_0=\frac{1}{2}$ and $S_N$ is a decreasing sequence by Karamata's inequality ($f(x)=\frac{1}{1+3x}$ is a convex function on $[0,1]$). A simpler alternative is given by comparing $S_N$ with $S_{N+3}$ then checking that $S_{N+3}\leq S_N$ and $S_1,S_2\leq\frac{1}{2}$.

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HINT: Let $f(t)$ be the sum. Show that $f(t+3) \leq f(t)$.

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Why not the integral test? We have $$\sum_{i=0}^{t-2}\frac{1}{t+3i}\leq\frac{1}{4t-6}+\int_{0}^{t-2}\frac{1}{t+3x}dx $$ $$=\frac{1}{4t-6}+\frac{1}{3}\log\left(4-\frac{6}{t}\right) $$ and now it is simple to prove that $f\left(t\right)=\frac{1}{4t-6}+\frac{1}{3}\log\left(4-\frac{6}{t}\right) $ is a monotonic increasing function if $t\geq3 $ so

$$\color{red}{f\left(t\right)\leq\lim_{t\rightarrow\infty}f\left(t\right)=\frac{\log\left(4\right)}{3}\approx0.462098.}$$

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