<s>[INST] Proving that $\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{100}} < 20$ How am I suppose to prove that:
$$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{100}} < 20$$
Do I use the way like how we count $1+2+ \cdots+100$ to estimate?
So $1/5050 \lt 20$, implying that it is indeed less than $20$?
 [/INST] We can estimate the sum by integration: because the inverse square root function is strictly descreasing, hence for $n\in \mathbb{N}$
$$
\frac{1}{\sqrt{n}}< \frac{1}{\sqrt{x}} \quad \text{ for } x\in (n-1,n).
$$
Integrating both sides on this interval:
$$
\frac{1}{\sqrt{n}}= \int^n_{n-1}\frac{1}{\sqrt{n}}\,dx< \int^{n}_{n-1}\frac{1}{\sqrt{x}}\,dx.
$$
Therefore
$$
\frac{1}{\sqrt{1}} < \int_0^1 \frac{1}{\sqrt{x}}\,dx
\\
\frac{1}{\sqrt{2}} < \int_1^2 \frac{1}{\sqrt{x}}\,dx
\\
\cdots
\\
\frac{1}{\sqrt{100}} < \int_{99}^{100} \frac{1}{\sqrt{x}}\,dx
$$
and
$$
\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + ... + \frac{1}{\sqrt{100}} < \int_0^{100}\frac{1}{\sqrt{x}}\,dx =20.
$$
</s>