Can we prove this inequality $1+\sqrt{2}+\cdots+\sqrt{n}<\frac{4n+3}{6}\sqrt{n}$ using integrals by parts?

Question

Can we prove this inequality $1+\sqrt{2}+\cdots+\sqrt{n}<\dfrac{4n+3}{6}\sqrt{n}$ using integrals by parts?

I need to prove this inequality, and I tried this way:

$$\begin{split} \int_0^n\sqrt{x} \,dx &=\sum\limits_{k=1}^n \int_{k-1}^k \sqrt{x}\, dx\\ &=\sum\limits_{k=1}^n \int_0^1 \sqrt{t+k-1} \, dt\\ &=\sum\limits_{k=1}^n \sqrt{k}-\sum\limits_{k=1}^n \int_0^1\dfrac{t}{2\sqrt{t+k-1}}\,dt \end{split}$$

Thus $$\sum\limits_{k=1}^n \sqrt{k}=\int_0^n \sqrt{x}\, dx+\sum\limits_{k=1}^n \int_0^1 \dfrac{t}{2\sqrt{t+k-1}}\, dt$$.

But maybe I didn't find some inequality trick, I cannot find a proper approxmation of the sum of the integrals.

Answer 1

First, see that we have

$$ \int_0^n \sqrt{x} \;dx = \frac{4n}{6}\sqrt{n} $$

which gives us a "part" of the demanded result. Moreover, it suggests that we should try to prove the following:

$$ \sum_{k=1}^n \int_0^1 \frac{t}{2\sqrt{t+k-1}} \;dt < \frac{3}{6}\sqrt{n} = \frac{1}{2}\sqrt{n} $$

Evaluating the integral we get:

\begin{align} \int_0^1 \frac{t}{2\sqrt{t+k-1}} \;dt &= \int_{k-1}^{k} \frac{x-k+1}{2\sqrt{x}} \;dx = \frac{1}{2} \int_{k-1}^{k} \sqrt{x} \;dx - (k-1) \int_{k-1}^{k} \frac{1}{2\sqrt{x}} \;dx \\ &= \frac{1}{3} \left(-2k\sqrt{k} + 2k\sqrt{k-1} + 3\sqrt{k} - 2\sqrt{k-1}\right) \end{align}

However, for all $k \geq 1$, you can prove that:

$$ \frac{1}{3} \left(-2k\sqrt{k} + 2k\sqrt{k-1} + 3\sqrt{k} - 2\sqrt{k-1}\right) < \frac{1}{2} \left(\sqrt{k}-\sqrt{k-1}\right) $$

So we end up with a telescoping sum and the result we wanted:

$$ \sum_{k=1}^n \int_0^1 \frac{t}{2\sqrt{t+k-1}} \;dt < \frac{1}{2} \sum_{k=1}^n \left(\sqrt{k}-\sqrt{k-1}\right) = \frac{1}{2} \sqrt{n} $$

Answer 2

Hint: Prove that $$\int_{k-1}^k\sqrt{x}dx>\dfrac{\sqrt{k}+\sqrt{k-1}}{2}$$ and then sum up to get your desired conclusion. You can just do this manually or generally this follows from a very well-known fact that the trapeziodal rule underestimates integral if the function is concave down. ($f(x) = \sqrt{x}$) is concave down)