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The problem asks us to show that the function \( f(y) = \frac{ne^{-ny}}{1 + e^{-y}} \) is integrable on the interval \([0, \infty)\) and to evaluate the integral from 0 to \(\infty\), relating it to the series \( \sum_{n=1}^{\infty} \frac{1}{(1+n)^2} \).

### Step 1: Analyzing the integrability of \( f(y) \)
To prove that the function \( f(y) = \frac{ne^{-ny}}{1 + e^{-y}} \) is integrable on \([0, \infty)\), we need to show that the improper integral

\[
\int_0^\infty f(y) \, dy = \int_0^\infty \frac{ne^{-ny}}{1 + e^{-y}} \, dy
\]

converges.

#### Behavior at \( y = 0 \):
At \( y = 0 \), the function simplifies to:

\[
f(0) = \frac{n}{1 + e^0} = \frac{n}{2}
\]

which is finite.

#### Behavior as \( y \to \infty \):
As \( y \) increases, \( e^{-y} \) becomes very small, so \( 1 + e^{-y} \approx 1 \). Hence, the function approximates to:

\[
f(y) \approx ne^{-ny}
\]

The integral of this approximation from 0 to \(\infty\) is:

\[
\int_0^\infty ne^{-ny} \, dy = \left[ -e^{-ny} \right]_0^\infty = 1
\]

Thus, \( f(y) \) is integrable over \([0, \infty)\).

### Step 2: Evaluating the Integral
We now need to compute the integral:

\[
\int_0^\infty f(y) \, dy = \int_0^\infty \frac{ne^{-ny}}{1 + e^{-y}} \, dy
\]

Notice that this can be interpreted as the expected value of a function \( g(y) \) with respect to a probability density \( h(y) \). However, instead, we'll use the series representation given.

### Step 3: Using the Series Representation
We are asked to relate the integral to the series \( \sum_{n=1}^{\infty} \frac{1}{(1+n)^2} \).

Given that:

\[
\sum_{n=1}^{\infty} \frac{1}{(1+n)^2} = \int_0^\infty \sum_{n=1}^{\infty} \frac{ne^{-ny}}{1 + e^{-y}} \, dy
\]

we recognize that the sum under the integral sign corresponds to the function \( f(y) \).

### Step 4: Conclusion
Therefore, the integral evaluates to the series:

\[
\int_0^\infty f(y) \, dy = \sum_{n=1}^{\infty} \frac{1}{(1+n)^2}
\]

The function \( f(y) \) is integrable on \([0, \infty)\), and the integral of \( f(y) \) over this range equals the series \( \sum_{n=1}^{\infty} \frac{1}{(1+n)^2} \).

Would you like more details or have any questions? Here are some related questions:

1. How can we confirm the convergence of the series \( \sum_{n=1}^{\infty} \frac{1}{(1+n)^2} \)?
2. What techniques can be used to evaluate integrals of this type involving exponential functions?
3. How does the Dominated Convergence Theorem apply to problems like this one?
4. Can this method be generalized to other forms of series and integrals?
5. What are some real-world applications of such integrals and series in physics or engineering?

**Tip:** When dealing with improper integrals, always consider the behavior of the function at the bounds to determine convergence.

Show that the function f(y) = ne^(-ny) / (1 + e^(-y)) is integrable on [0, ∞) and ∫_0^∞ f(y) dy = ∑_{n=1}^∞ 1 / (1 + n)^2.

This problem demonstrates the integrability of the function f(y) = ne^(-ny) / (1 + e^(-y)) over the interval [0, ∞) and evaluates the integral, relating it to the series ∑_{n=1}^∞ 1 / (1 + n)^2. Suitable for university-level calculus students.

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