Disclaimer: This requires calculus.
Integrals:
$$\sum_{k=1}^n\frac1{\sqrt n}\ge\int_1^{n+1}\frac1{\sqrt x}\ dx=2\sqrt{n+1}-2$$
And it's very easy to check that
$$2\sqrt{n+1}-2\ge\sqrt n$$
for $n\ge2$.
A visuallization of this argument:
From the red lines down, that area represents a sum. From the blue line down, that represents an integral. Clearly, the integral is smaller than the sum.
Disclaimer: This requires calculus.
Integrals:
$$\sum_{k=1}^n\frac1{\sqrt n}\ge\int_1^{n+1}\frac1{\sqrt x}\ dx=2\sqrt{n+1}-2$$
And it's very easy to check that
$$2\sqrt{n+1}-2\ge\sqrt n$$
for $n\ge2$.
A visuallization of this argument:
From the red lines down, that area represents a sum. From the blue line down, that represents an integral. Clearly, the integral is smaller than the sum.
Integrals:
$$\sum_{k=1}^n\frac1{\sqrt n}\ge\int_1^{n+1}\frac1{\sqrt x}\ dx=2\sqrt{n+1}-2$$
And it's very easy to check that
$$2\sqrt{n+1}-2\ge\sqrt n$$
for $n\ge2$.
A visuallization of this argument:
From the red lines down, that area represents a sum. From the blue line down, that represents an integral. Clearly, the integral is smaller than the sum.
Disclaimer: This requires calculus.
Integrals:
$$\sum_{k=1}^n\frac1{\sqrt n}\ge\int_1^n\frac1{\sqrt x}\ dx=2\sqrt n-2$$$$\sum_{k=1}^n\frac1{\sqrt n}\ge\int_1^{n+1}\frac1{\sqrt x}\ dx=2\sqrt{n+1}-2$$
And it's very easy to check that
$$2\sqrt n-2\ge\sqrt n$$$$2\sqrt{n+1}-2\ge\sqrt n$$
for $n\ge2$.
A visuallization of this argument:
From the red lines down, that area represents a sum. From the blue line down, that represents an integral. Clearly, the integral is smaller than the sum.
Disclaimer: This requires calculus.
Integrals:
$$\sum_{k=1}^n\frac1{\sqrt n}\ge\int_1^n\frac1{\sqrt x}\ dx=2\sqrt n-2$$
And it's very easy to check that
$$2\sqrt n-2\ge\sqrt n$$
for $n\ge2$.
A visuallization of this argument:
From the red lines down, that area represents a sum. From the blue line down, that represents an integral. Clearly, the integral is smaller than the sum.
Disclaimer: This requires calculus.
Integrals:
$$\sum_{k=1}^n\frac1{\sqrt n}\ge\int_1^{n+1}\frac1{\sqrt x}\ dx=2\sqrt{n+1}-2$$
And it's very easy to check that
$$2\sqrt{n+1}-2\ge\sqrt n$$
for $n\ge2$.
A visuallization of this argument:
From the red lines down, that area represents a sum. From the blue line down, that represents an integral. Clearly, the integral is smaller than the sum.
Disclaimer: This requires calculus.
Integrals:
$$\sum_{k=1}^n\frac1{\sqrt n}\ge\int_1^n\frac1{\sqrt x}\ dx=2\sqrt n-2$$
And it's very easy to check that
$$2\sqrt n-2\ge\sqrt n$$
for $n\ge2$.
A visuallization of this argument:
From the red lines down, that area represents a sum. From the blue line down, that represents an integral. Clearly, the integral is smaller than the sum.
Integrals:
$$\sum_{k=1}^n\frac1{\sqrt n}\ge\int_1^n\frac1{\sqrt x}\ dx=2\sqrt n-2$$
And it's very easy to check that
$$2\sqrt n-2\ge\sqrt n$$
for $n\ge2$.
Disclaimer: This requires calculus.
Integrals:
$$\sum_{k=1}^n\frac1{\sqrt n}\ge\int_1^n\frac1{\sqrt x}\ dx=2\sqrt n-2$$
And it's very easy to check that
$$2\sqrt n-2\ge\sqrt n$$
for $n\ge2$.
A visuallization of this argument:
From the red lines down, that area represents a sum. From the blue line down, that represents an integral. Clearly, the integral is smaller than the sum.
