The sequence of function $f_k(x) = \frac{x^k}{(1+x)^2},~x\ge 0$

Question

For non-negative integers $k\geq 1$ define $$ f_k(x) = \frac{x^k}{(1+x)^2},~x\geq 0. $$ Which of the following statements are true?

1. For each $k$, $f_k$ is a function of bounded variation on compact intervals
2. For every $k$, $\int_0^1 f_k(x)~\mathrm{d} x < \infty $
3. $\lim_{k\to \infty } \int_0^1 f_k(x) \mathrm{d} x$ exists
4. The sequence of functions $f_k$ converge uniformly on $[0,1]$ as $k\to \infty $

Note that in (4), we have $$f(x) = \begin{cases}0,~\text{ if }x<1 \\ 1/4,~\text{ if }x=1,\end{cases}$$ which is not continuous, and hence the convergence is not uniform.

For option (2), each of the functions is continuous and hence bounded on $[0,1]$ and therefore, the integral $\int_0^1 f_k(x)~\mathrm{d}x$ will be finite.

I got stuck with options (1) and (3). Kindly help me!

Answer 1

1. Any function with a bounded derivative is of bounded variation: Apply Mean Value Theorem to prove this. Each $f_k$ has a bounded derivative.

2. $0 \leq f_k(x)\leq \frac 1 {(1+x)^{2}}$. By Dominated Convergence Theorem we see that $\lim \int_0^{1} f_k(x)dx=\int_0^{1} f(x)dx=0$.