Given the series $$ \sum_{n=1}^{\infty} \frac{n}{1+n^{2}} $$ use the Integral Test to determine if it converges or diverges.

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We start by defining the function
$$
f(x)=\frac{x}{1+x^2}
$$
which is positive, continuous, and eventually decreasing for $$ x\geq1 $$.

**Step 1. Verify that $$ f(x) $$ is decreasing:**

Compute the derivative:
$$
f'(x)=\frac{d}{dx}\left(\frac{x}{1+x^2}\right)=\frac{(1+x^2)(1)-x(2x)}{(1+x^2)^2} = \frac{1+x^2-2x^2}{(1+x^2)^2}=\frac{1-x^2}{(1+x^2)^2}.
$$
For $$ x>1 $$, the numerator $$ 1-x^2 $$ is negative, so $$ f'(x) <0 $$. Hence, $$ f(x) $$ is decreasing for $$ x\geq 1 $$.

**Step 2. Set up the Integral Test:**

The Integral Test states that the convergence of the series
$$
\sum_{n=1}^{\infty} \frac{n}{1+n^2}
$$
can be determined by studying the improper integral
$$
\int_{1}^{\infty} \frac{x}{1+x^2}\,dx.
$$

**Step 3. Evaluate the integral:**

Let
$$
I=\int_{1}^{\infty} \frac{x}{1+x^2}\,dx.
$$
Use the substitution
$$
u=1+x^2 \quad \text{so that} \quad du=2x\,dx \quad \text{or} \quad x\,dx=\frac{1}{2}\,du.
$$
Changing the limits:
- When $$ x=1 $$, $$ u=1+1^2=2 $$,
- When $$ x\to \infty $$, $$ u\to \infty $$.

Thus, the integral becomes:
$$
I=\int_{u=2}^{\infty} \frac{1}{2}\cdot\frac{1}{u}\,du=\frac{1}{2}\int_{2}^{\infty}\frac{1}{u}\,du.
$$

Now evaluate the integral:
$$
\frac{1}{2}\int_{2}^{\infty}\frac{1}{u}\,du=\frac{1}{2}\left[\ln|u|\right]_{2}^{\infty}=\frac{1}{2}\lim_{b\to\infty} (\ln b - \ln2).
$$
Since $$\lim_{b\to\infty}\ln b = \infty$$, the integral diverges:
$$
I=\infty.
$$

**Step 4. Conclusion via the Integral Test:**

Because the integral
$$
\int_{1}^{\infty}\frac{x}{1+x^2}\,dx
$$
diverges, by the Integral Test, the series
$$
\sum_{n=1}^{\infty} \frac{n}{1+n^2}
$$
also diverges.
The series $$ \sum_{n=1}^{\infty} \frac{n}{1+n^{2}} $$ diverges.

Given the series

use the Integral Test to determine if it converges or diverges.

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Given the series use the Integral Test to determine if it converges or diverges.