integration 5

$\begin{aligned} I & = \int_0^2 \left(\lim_{n \to \infty} \sum_{k=1}^\infty \frac n{(n+k)^2} + \sum_{k=1}^\infty \blue{\lim_{n \to \infty} \frac n{(n+k)^2}} \right) dx & \small \blue{\text{A }\infty/\infty \text{ case, L'Hôpital's rule applies.}} \\ & = \int_0^2 \left(\red{\lim_{n \to \infty} \sum_{k=1}^\infty \frac 1{n\left(1+\frac kn \right)^2}} + \sum_{k=1}^\infty \blue{\lim_{n \to \infty} \frac 1{2(n+k)}} \right) dx & \small \blue{\text{Differentiate up and down w.r.t. }n} \\ & = \int_0^2 \left(\red{\int_0^1 \frac 1{(1+x)^2} dx} + \sum_{k=1}^\infty \blue{0} \right) dx & \small \red{\text{By Riemann sums}} \\ & = \int_0^2 \red{\left[\frac 1{1+x} \right]_1^0} dx = \int_0^2 \frac 12 \ dx = \frac x2 \ \bigg|_0^2 = \boxed 1 \end{aligned}$

References:

• L'Hôpital's rule
• Riemann sums

Consider the sum:

$\lim_{n \to \infty} \sum_{k=1}^{n} \frac{n}{(n+k)^2} = S_1$

$\implies \lim_{n \to \infty}\sum_{k=1}^{n} \frac{\frac{1}{n}}{\left(1+\frac{k}{n}\right)^2}=S_1$

Take $1/n = dx$ and $k/n=x$ This transforms the sum to the following integral as n tends to infinity:

$S_1 = \int_{0}^{1}\frac{dx}{(1+x)^2}=\frac{1}{2}$

Again, consider the sum:

$\sum_{k=1}^{\infty} \lim_{n \to \infty} \frac{n}{(n+k)^2} = S_2$

$S_2 = \sum_{k=1}^{\infty} \left(\lim_{n \to \infty} \frac{\frac{1}{n}}{\left(1+\frac{k}{n}\right)^2}\right)$ $S_2 =\sum_{k=1}^{\infty} 0$ $\because \lim_{n \to \infty} \frac{\frac{1}{n}}{\left(1+\frac{k}{n}\right)^2}=0$ $\implies S_2 = 0$

Therefore:

$\int_{0}^{2}\left(S_1 + S_2\right)dx=1$

See the upper limit for the sum definition that I have used for $S_1$ . I think that is the correct notation rather than what is used in the problem statement.