Calculate the upper boundary of the given sum

Question

How to find upper boundary of $\sum^{\infty}_{k = 1}\frac{1}{k^{3/2}}$? I know that it can be solved using integrals, but can someone explain how this sum is connected with integral of $f(x) = \frac{1}{k^{3/2}}$?

Answer 1

One may recall that, for any non-negative continuous function $f$ defined on $[N,\infty)$ on which it is monotone decreasing, we have $$ \int_N^\infty f(x)\,dx\le\sum_{k=N}^\infty f(k)\le f(N)+\int_N^\infty f(x)\,dx,\qquad N>0, $$ giving here, with $f(x):=\dfrac{1}{x^{3/2}}$ and $N=1$, $$ \int_1^\infty \dfrac{1}{x^{3/2}}\,dx<\sum_{k=1}^\infty \dfrac{1}{k^{3/2}}< \dfrac{1}{1^{3/2}}+\int_1^\infty \dfrac{1}{x^{3/2}}\,dx $$ that is

$$ 2<\sum_{k=1}^\infty \dfrac{1}{k^{3/2}}< 3. $$

Remark. In fact $$ \sum_{k=1}^\infty \dfrac{1}{k^{3/2}}=\zeta\left(\frac32 \right)=2.612375348685488343348567564312 \cdots. $$