1. Given sums

We have

$$S_n=\sum_{k=1}^n \frac{n}{n^2+kn+k^2}, \qquad T_n=\sum_{k=0}^{n-1} \frac{n}{n^2+kn+k^2}.$$

We need to compare them with

$$\frac{\pi}{3\sqrt{3}}.$$

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2. Rewrite the general term

Factor out $n^2$ from the denominator:

$$\frac{n}{n^2+kn+k^2} =\frac{n}{n^2\left(1+\frac{k}{n}+\frac{k^2}{n^2}\right)} =\frac{1}{n}\cdot \frac{1}{1+\frac{k}{n}+\left(\frac{k}{n}\right)^2}.$$

Define

$$f(x)=\frac{1}{1+x+x^2}.$$

Then

$$S_n=\sum_{k=1}^n \frac{1}{n}f\left(\frac{k}{n}\right), \qquad T_n=\sum_{k=0}^{n-1} \frac{1}{n}f\left(\frac{k}{n}\right).$$

So both are Riemann sums for $\int_0^1 f(x)\,dx$.

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3. Monotonicity of $f(x)$ on $[0,1]$

Compute derivative:

$$f'(x)= -\frac{1+2x}{(1+x+x^2)^2}.$$

For $x\in[0,1]$, we have $1+2x>0$, hence

$$f'(x)<0.$$

Thus $f$ is strictly decreasing on $[0,1]$.

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4. Interpret $S_n$ and $T_n$ as right and left Riemann sums

Since $f$ is decreasing:

• The right-endpoint sum is less than the integral.
• The left-endpoint sum is greater than the integral.

Now,

$$S_n=\sum_{k=1}^n \frac{1}{n}f\left(\frac{k}{n}\right)$$

is the right Riemann sum on $[0,1]$, so

$$S_n<\int_0^1 \frac{dx}{1+x+x^2}.$$

And

$$T_n=\sum_{k=0}^{n-1} \frac{1}{n}f\left(\frac{k}{n}\right)$$

is the left Riemann sum, so

$$T_n>\int_0^1 \frac{dx}{1+x+x^2}.$$

Therefore it remains to compute the integral.

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5. Evaluate the integral

We need

$$I=\int_0^1 \frac{dx}{1+x+x^2}.$$

Complete the square:

$$x^2+x+1=\left(x+\frac12\right)^2+\frac34.$$

So

$$I=\int_0^1 \frac{dx}{\left(x+\frac12\right)^2+\left(\frac{\sqrt3}{2}\right)^2}.$$

Using

$$\int \frac{dx}{u^2+a^2}=\frac{1}{a}\tan^{-1}\left(\frac{u}{a}\right),$$

we get

$$I=\frac{2}{\sqrt3}\left[\tan^{-1}\left(\frac{2x+1}{\sqrt3}\right)\right]_0^1.$$

Thus

$$I=\frac{2}{\sqrt3}\left(\tan^{-1}(\sqrt3)-\tan^{-1}\left(\frac{1}{\sqrt3}\right)\right).$$

Now,

$$\tan^{-1}(\sqrt3)=\frac{\pi}{3}, \qquad \tan^{-1}\left(\frac1{\sqrt3}\right)=\frac{\pi}{6}.$$

Hence

$$I=\frac{2}{\sqrt3}\left(\frac{\pi}{3}-\frac{\pi}{6}\right) =\frac{2}{\sqrt3}\cdot \frac{\pi}{6} =\frac{\pi}{3\sqrt3}.$$

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6. Final comparison

From Steps 4 and 5:

$$S_n<\int_0^1 \frac{dx}{1+x+x^2}=\frac{\pi}{3\sqrt3},$$

and

$$T_n>\int_0^1 \frac{dx}{1+x+x^2}=\frac{\pi}{3\sqrt3}.$$

So:

• A is correct
• B is incorrect
• C is incorrect
• D is correct

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7. Comparison with stored answer

Stored correct answer: A, D

Our derived answer: A, D

They match.