Exercises on limits of integrals and notions of convergence

The exercises compute limits and series of integrals of functions on intervals of $\mathbb{R}$ with the Lebesgue measure ${\lambda,}$ and study the convergence of sequences of such functions.

Exercises#

A sequence with $\tanh(nx)$#

Remark 1. For every ${y \in \mathbb{R},}$ the hyperbolic sine and the hyperbolic cosine of $y$ are

$$\sinh(y) = \frac{e^y - e^{-y}}{2}, \qquad \cosh(y) = \frac{e^y + e^{-y}}{2},$$

and ${\tanh(y) = \sinh(y)/\cosh(y).}$ Figure 1 draws the graph of the function ${\tanh,}$ with the dashed levels 1 and ${-1.}$

Figure 1. The drawing of Remark 1: the graph of the function $\tanh$ on ${\mathbb{R},}$ with the dashed levels 1 and ${-1.}$

Example 2. Compute ${\lim_n \int_0^2 f_n \, d\lambda,}$ where, for every integer ${n \ge 1,}$ the function ${f_n \colon (0, 2) \to \mathbb{R}}$ is given by

$$f_n(x) = \frac{\tanh(nx)}{\sqrt{\sin x}} \, \frac{1}{2 + n x^3}.$$

The functions $f_n$ of Example 2 are nonnegative on ${(0, 2).}$ For every ${x \in (0, 2),}$ since ${\tanh(nx) \le 1,}$

$$\begin{aligned} 0 \le f_n(x) = |f_n(x)| &= \left| \frac{\tanh(nx)}{\sqrt{\sin x}} \, \frac{1}{2 + n x^3} \right| \\ &\le \frac{1}{\sqrt{\sin x}} \, \frac{1}{2 + n x^3} \\ &= \frac{1}{n} \, \frac{1}{\sqrt{\sin x}} \, \frac{1}{\frac{2}{n} + x^3} \xrightarrow[n \to \infty]{} 0, \end{aligned}$$

because ${2/n \to 0.}$ Therefore ${f_n \to 0}$ pointwise on ${(0, 2).}$ Moreover, since ${\tanh(nx) \le 1}$ and ${2 + n x^3 \ge 2,}$

$$0 \le f_n(x) = \frac{\tanh(nx)}{\sqrt{\sin x}} \, \frac{1}{2 + n x^3} \le \frac{1}{2 \sqrt{\sin x}} =: g(x).$$

We check that ${\int_0^2 1/\sqrt{\sin x} \, d\lambda < +\infty,}$ with the graph of $\sin x$ in Figure 2.

Figure 2. The drawing of Example 2: the graph of $\sin x$ for ${x \in [0, 2],}$ with its maximum 1 at ${\pi/2,}$ the dotted segment at ${x = 2,}$ and the points ${\pi/2,}$ 2 and $\pi$ on the axis.

As ${x \to 0^+}$ we have ${\sin x \sim x,}$ hence ${1/\sqrt{\sin x} \sim 1/\sqrt{x}.}$ Moreover ${\int_0^2 1/\sqrt{x} \, d\lambda < +\infty,}$ hence ${g \in L^1((0, 2)).}$ By the dominated convergence theorem, Theorem 8 of the previous post with ${p = 1,}$ ${f_n \to 0}$ in ${L^1((0, 2)).}$

Remark 3. Let ${(X, \mathcal{M}, \mu)}$ be a measure space and let ${f_n, f \in L^1(X)}$ for every ${n.}$ If ${f_n \to f}$ in ${L^1(X),}$ then

$$\begin{aligned} \left| \int_X f_n \, d\mu - \int_X f \, d\mu \right| &= \left| \int_X (f_n - f) \, d\mu \right| \\ &\le \int_X |f_n - f| \, d\mu = \|f_n - f\|_{L^1} \xrightarrow[n \to \infty]{} 0. \end{aligned}$$

By Remark 3, applied to the functions $f_n$ of Example 2 with ${f = 0,}$

$$\lim_n \int_0^2 f_n \, d\lambda = \int_0^2 0 \, d\lambda = 0.$$

A sequence with $\arctan(nx)$#

Example 4. Compute ${\lim_n \int_0^\infty f_n \, d\lambda,}$ where, for every integer ${n \ge 1,}$ the function ${f_n \colon (0, +\infty) \to \mathbb{R}}$ is given by

$$f_n(x) = \frac{\dfrac{\pi}{2} - \arctan(nx)}{1 + x}.$$

Remark 5. Figure 3 draws the graph of the function arctan on ${\mathbb{R},}$ with the dashed level $\pi/2$ and the value $-\pi/2$ marked on the left.

Figure 3. The drawing of Remark 5: the graph of the function arctan on ${\mathbb{R},}$ with the dashed level $\pi/2$ and the value $-\pi/2$ marked on the left.

Since ${\arctan(nx) < \pi/2}$ for every ${x \in (0, \infty),}$ the functions $f_n$ of Example 4 are positive on ${(0, \infty).}$ Let ${x \in (0, +\infty).}$ Then ${nx \to +\infty}$ as ${n \to +\infty,}$ hence ${\lim_{n \to \infty} \arctan(nx) = \pi/2.}$ Therefore ${f_n(x) = (\pi/2 - \arctan(nx))/(1 + x) \to 0}$ as ${n \to +\infty,}$ and ${f_n \to 0}$ pointwise on ${(0, \infty).}$ Observe that, since ${-\arctan(nx) \le \pi/2,}$

$$\begin{aligned} |f_n| = \left| \frac{\dfrac{\pi}{2} - \arctan(nx)}{1 + x} \right| &= \frac{\dfrac{\pi}{2} - \arctan(nx)}{1 + x} \\ &\le \frac{\pi}{1 + x} \underset{x \to \infty}{\sim} \frac{\pi}{x}, \end{aligned}$$

and this bound does not belong to ${L^1((1, \infty)).}$ Now we analyze ${\pi/2 - \arctan(nx){:}}$

$$\begin{aligned} \frac{\pi}{2} - \arctan(nx) &= \frac{\pi}{2} - \int_0^{nx} \arctan'(t) \, dt \\ &= \int_0^\infty \arctan'(t) \, dt - \int_0^{nx} \arctan'(t) \, dt \\ &= \int_{nx}^\infty \arctan'(t) \, dt = \int_{nx}^\infty \frac{1}{1 + t^2} \, dt \\ &\le \int_{nx}^\infty \frac{1}{t^2} \, dt = \frac{1}{nx}, \end{aligned}$$

where ${1/(1 + t^2) \le 1/t^2}$ for every ${t > 0.}$ The last integral is computed as a limit:

$$\int_{nx}^\infty \frac{1}{t^2} \, dt = \lim_{R \to \infty} \int_{nx}^R \frac{1}{t^2} \, dt = \lim_{R \to \infty} \Big( -\frac{1}{t} \, \Big|_{nx}^R \Big) = \lim_{R \to \infty} \Big( -\frac{1}{R} + \frac{1}{nx} \Big).$$

Hence, dividing the chain by ${1 + x,}$

$$\begin{aligned} |f_n| = f_n = \frac{\dfrac{\pi}{2} - \arctan(nx)}{1 + x} &\le \frac{\dfrac{1}{nx}}{1 + x} \\ &= \frac{1}{n} \, \frac{1}{x (1 + x)} \le \frac{1}{x (1 + x)} =: g_1(x), \end{aligned}$$

because ${1/n \le 1,}$ and ${g_1 \in L^1((1, \infty)).}$ Notice that ${g_1 \notin L^1((0, 1)).}$ Also ${|f_n(x)| \le \pi/(1 + x) =: g_2(x),}$ and ${g_2 \in L^1((0, 1)).}$ Therefore

$$|f_n| \le g_2 \chi_{(0, 1)} + g_1 \chi_{(1, \infty)} \qquad \text{almost everywhere on } (0, +\infty),$$

where ${g_2 \chi_{(0, 1)}}$ and ${g_1 \chi_{(1, \infty)}}$ belong to ${L^1((0, \infty)),}$ and so does their sum. By the dominated convergence theorem, Theorem 8 of the previous post with ${p = 1,}$ ${f_n \to 0}$ in ${L^1((0, \infty)).}$ Hence, by Remark 3,

$$\lim_n \int_0^\infty f_n \, d\lambda = \int_0^\infty 0 \, d\lambda = 0.$$

Remark 6. For the main estimate of Example 4, the one on ${\pi/2 - \arctan(nx),}$ one may also use the identity

$$\arctan(y) + \arctan\Big( \frac{1}{y} \Big) = \frac{\pi}{2} \qquad \text{for every } y \in (0, \infty).$$

Applied with ${y = nx,}$ it gives, by Taylor’s formula for arctan at 0, since ${1/(nx) \to 0}$ as ${n \to \infty,}$

$$\frac{\pi}{2} - \arctan(nx) = \arctan\Big( \frac{1}{nx} \Big) = \frac{1}{nx} + o\Big( \frac{1}{nx} \Big) \qquad \text{as } n \to \infty.$$

A series of integrals#

Example 7. Compute

$$\sum_{n=1}^\infty \int_0^1 x^n (1 - x) e^x \, d\lambda.$$

In the series of Example 7, for every ${x \in (0, 1)}$ the factors ${x^n,}$ ${1 - x}$ and $e^x$ are positive, so ${x^n (1 - x) e^x \ge 0}$ on ${(0, 1).}$ Hence, by the monotone convergence theorem, in the form of Remark 7 of the sixth post,

$$\begin{aligned} &\sum_{n=1}^\infty \int_0^1 x^n (1 - x) e^x \, d\lambda \\ &\quad = \int_0^1 \sum_{n=1}^\infty x^n (1 - x) e^x \, d\lambda. \end{aligned}$$

Remark 8. For every ${q \in \mathbb{R}}$ with ${|q| < 1,}$

$$\sum_{n=0}^\infty q^n = \frac{1}{1 - q}.$$

By Remark 8, applied with ${q = x,}$ the series inside the last integral can be summed for every ${x \in (0, 1){:}}$

$$\begin{aligned} \sum_{n=1}^\infty x^n (1 - x) e^x &= (1 - x) e^x \sum_{n=1}^\infty x^n \\ &= (1 - x) e^x \Biggl( \biggl( \sum_{n=0}^\infty x^n \biggr) - 1 \Biggr) \\ &= (1 - x) e^x \biggl( \frac{1}{1 - x} - 1 \biggr) \\ &= (1 - x) e^x \, \frac{1 - 1 + x}{1 - x} = x e^x. \end{aligned}$$

Therefore the series of Example 7 equals ${\int_0^1 x e^x \, d\lambda = 1.}$

A sequence with exponentials#

Example 9. On ${[0, +\infty)}$ with the Lebesgue measure ${\lambda,}$ for every integer ${n \ge 1}$ let ${f_n \colon [0, +\infty) \to \mathbb{R}}$ be given by

$$f_n(x) = \frac{e^{-nx} + n^2 e^{-x}}{n + n^2 + (1 + n^2) x^2}.$$

(1) Study the convergence almost everywhere of ${(f_n).}$

For the functions $f_n$ of Example 9 and every ${x \in [0, +\infty),}$ dividing the numerator and the denominator by ${n^2,}$

$$\begin{aligned} f_n(x) &= \frac{n^{-2} e^{-nx} + e^{-x}}{\frac{1}{n} + 1 + \frac{1 + n^2}{n^2} \, x^2} \\ &\xrightarrow[n \to +\infty]{} \frac{e^{-x}}{1 + x^2}, \end{aligned}$$

since $n^{-2}$ and $1/n$ tend to ${0,}$ ${e^{-nx} \le 1}$ and ${(1 + n^2)/n^2 \to 1.}$ Hence ${f_n \to f = e^{-x}/(1 + x^2)}$ pointwise on ${[0, +\infty),}$ and therefore ${f_n \to f}$ almost everywhere on ${[0, +\infty).}$

(2) Study the convergence of $(f_n)$ in ${L^1((0, \infty)).}$ Notice that ${f_n \ge 0.}$ Hence, since $n^{-2}$ and $e^{-nx}$ are at most ${1,}$ $1/n$ is nonnegative and ${(1 + n^2)/n^2 \ge 1,}$

$$\begin{aligned} |f_n| = f_n &= \frac{n^{-2} e^{-nx} + e^{-x}}{\frac{1}{n} + 1 + \frac{1 + n^2}{n^2} \, x^2} \\ &\le \frac{1 + e^{-x}}{1 + x^2} =: g(x). \end{aligned}$$

Notice that ${g(x) \sim 1/x^2}$ as ${x \to \infty,}$ so that ${g \in L^1((1, \infty)).}$ Also ${\int_0^1 g \, d\lambda < +\infty,}$ hence ${g \in L^1((0, \infty)).}$ By the dominated convergence theorem, Theorem 8 of the previous post with ${p = 1,}$ ${f_n \to f}$ in ${L^1((0, \infty)).}$

(3) Let ${g_n(x) = f_n(x) + \chi_{[n, n+1]}(x).}$ Study the convergence of $(g_n)$ in ${L^1((0, \infty)).}$ We have ${f_n \to f}$ almost everywhere on ${(0, \infty)}$ and ${\chi_{[n, n+1]} \to 0}$ almost everywhere on ${(0, \infty),}$ hence ${g_n \to f}$ almost everywhere on ${(0, \infty).}$ Observe that, by the uniqueness of limits, Remark 12 of the previous post, if $(g_n)$ converges in ${L^1,}$ the limit must be ${f.}$ So we test the definition:

$$\int_0^\infty |g_n - f| \, d\lambda = \int_0^\infty |f_n + \chi_{[n, n+1]} - f| \, d\lambda.$$

Remark 10. For all ${a, b \in \mathbb{R},}$

$$|a + b| \ge \bigl|\, |a| - |b| \,\bigr| \ge |a| - |b|.$$

Applying Remark 10 at every ${x \in (0, \infty),}$ with ${a = \chi_{[n, n+1]}(x)}$ and ${b = f_n(x) - f(x),}$ and integrating over ${(0, \infty),}$

$$\begin{aligned} \int_0^\infty |g_n - f| \, d\lambda &\ge \int_0^\infty \big( \chi_{[n, n+1]} - |f_n - f| \big) \, d\lambda \\ &= 1 - \int_0^\infty |f_n - f| \, d\lambda. \end{aligned}$$

Hence ${\|g_n - f\|_{L^1} \ge 1 - \|f_n - f\|_{L^1},}$ where ${\lim_n \|f_n - f\|_{L^1} = 0}$ by (2). Taking the lower limit as ${n \to \infty,}$

$$\varliminf_n \|g_n - f\|_{L^1} \ge 1 - \lim_n \|f_n - f\|_{L^1} = 1.$$

Therefore ${\|g_n - f\|_{L^1} \nrightarrow 0}$ as ${n \to \infty,}$ and $(g_n)$ does not converge in ${L^1.}$

A sequence with a logarithm and the powers $(x/2)^n$#

Example 11. On ${(1, \infty)}$ with the Lebesgue measure ${\lambda,}$ for every integer ${n \ge 1}$ consider the functions ${f_n \ge 0}$ given by

$$f_n(x) = \frac{n^2 \log\big( 1 + \frac{2}{nx} \big)}{x + n}.$$

(1) Study the convergence almost everywhere and in $L^1$ of ${(f_n).}$

For the functions $f_n$ of Example 11, let ${x \in (1, \infty).}$ Since ${2/(nx) \to 0}$ as ${n \to \infty,}$ Taylor’s formula for the logarithm gives, as ${n \to \infty,}$

$$\begin{aligned} f_n(x) = \frac{n^2 \log\big( 1 + \frac{2}{nx} \big)}{n \big( \frac{x}{n} + 1 \big)} &= \frac{n}{\frac{x}{n} + 1} \biggl( \frac{2}{nx} + o\Big( \frac{2}{nx} \Big) \biggr) \\ &= \frac{n}{\frac{x}{n} + 1} \, \frac{2}{nx} \big( 1 + o(1) \big) \\ &= \frac{2}{\big( \frac{x}{n} + 1 \big) x} \big( 1 + o(1) \big), \end{aligned}$$

where ${x/n \to 0}$ and ${o(1) \to 0.}$ Hence ${f_n(x) \to 2/x}$ as ${n \to \infty,}$ for every ${x \in (1, \infty).}$ Therefore ${f_n \to f := 2/x}$ almost everywhere on ${(1, \infty).}$ By the uniqueness of limits, Remark 12 of the previous post, if $(f_n)$ converges in ${L^1((1, \infty))}$ to some function, that function must be ${f.}$

Remark 12. Let ${(X, \mathcal{M}, \mu)}$ be a measure space, let ${(f_n) \subset L^1(X)}$ and let ${f \colon X \to \mathbb{R}}$ be measurable. If ${f_n \to f}$ in ${L^1(X),}$ then ${f \in L^1(X).}$

By Remark 12, since ${f(x) = 2/x \notin L^1((1, \infty)),}$ ${f_n \nrightarrow f}$ in ${L^1((1, \infty)),}$ and therefore $(f_n)$ does not converge in ${L^1((1, \infty)).}$ Indeed, Remark 12 applies, because ${\log(1 + t) \le t}$ for ${t > 0}$ gives ${0 \le f_n(x) \le 2n/x^2,}$ so that each $f_n$ belongs to ${L^1((1, \infty)).}$

Part 2 of Example 11 concerns the functions ${g_n(x) = 2^{-n} x^n = (x/2)^n}$ for ${x \in \mathbb{R}}$ and every integer ${n \ge 1.}$ Figure 4 draws $g_n$ for $n$ even and for $n$ odd.

Figure 4. The first drawing of Part 2 of Example 11: the graph of $g_n$ for $n$ even and for $n$ odd, on a stretch of ${\mathbb{R}.}$ An instance with ${n = 2}$ and ${n = 3.}$

(1) Find the set ${D \subset \mathbb{R}}$ such that $(g_n)$ converges pointwise on ${D.}$ For every ${x \in \mathbb{R},}$

$$\Big( \frac{x}{2} \Big)^n \begin{cases} \xrightarrow[n \to \infty]{} 0 & \text{if } |x| < 2, \\ = 1^n = 1 \xrightarrow[n \to \infty]{} 1 & \text{if } x = 2, \\ = (-1)^n, \text{ which has no limit as } n \to \infty, & \text{if } x = -2, \\ \xrightarrow[n \to \infty]{} +\infty & \text{if } x > 2, \\ \text{has no limit as } n \to \infty & \text{if } x < -2. \end{cases}$$

Hence ${D = (-2, 2],}$ with

$$g_n(x) \xrightarrow[n \to \infty]{} \begin{cases} 0 & x \in (-2, 2), \\ 1 & x = 2. \end{cases}$$

(2) Study the convergence of $(g_n)$ in ${L^1(D).}$ Notice that ${g_n \to 0}$ almost everywhere on ${D.}$ Moreover, for every ${x \in D = (-2, 2],}$ since ${|x| \le 2,}$

$$|g_n(x)| = \left| \Big( \frac{x}{2} \Big)^n \right| = \frac{|x|^n}{2^n} \le \frac{2^n}{2^n} = 1.$$

The constant function ${h = 1}$ on $D$ satisfies ${h \in L^1(D).}$ By the dominated convergence theorem, Theorem 8 of the previous post with ${p = 1,}$ ${g_n \to 0}$ in ${L^1(D).}$

(3) Study the convergence of $(g_n)$ in ${L^\infty(D).}$ By the uniqueness of limits, Remark 12 of the previous post, if $(g_n)$ converges in ${L^\infty(D),}$ then it must converge to zero in ${L^\infty(D).}$ Suppose, by contradiction, that ${g_n \to 0}$ in ${L^\infty(D).}$ Then ${\|g_n\|_{L^\infty(D)} \to 0.}$ Figure 5 draws $g_n$ on $D$ for $n$ even and for $n$ odd, with ${g_n(2) = 1.}$

Figure 5. The second drawing of Part 2 of Example 11: the graph of $g_n$ on ${D = (-2, 2]}$ for $n$ even and for $n$ odd, with the value ${g_n(2) = 1.}$ The empty circles mark the point ${x = -2,}$ which is not in ${D.}$ An instance with ${n = 2}$ and ${n = 3.}$

Fix $n$ and ${\varepsilon > 0.}$ Because $g_n$ is continuous and ${g_n(2) = 1,}$ there exists ${\delta > 0}$ such that ${g_n \ge 1 - \varepsilon}$ on ${(2 - \delta, 2].}$ Since ${\lambda((2 - \delta, 2]) = \delta > 0,}$ it follows that ${\|g_n\|_{L^\infty(D)} \ge 1 - \varepsilon.}$ This is true for every ${\varepsilon > 0,}$ for any fixed ${n,}$ hence

$$\begin{aligned} &\|g_n\|_{L^\infty(D)} \ge 1 \quad \text{for every } n \\ &\Longrightarrow \quad \varliminf_n \|g_n\|_{L^\infty(D)} \ge 1 \\ &\Longrightarrow \quad \|g_n\|_{L^\infty(D)} \nrightarrow 0 \quad \text{as } n \to \infty. \end{aligned}$$

This contradicts ${\|g_n\|_{L^\infty(D)} \to 0.}$ Hence $(g_n)$ does not converge in ${L^\infty(D).}$

Remark 13. In (3) of Part 2 of Example 11, the norm of $g_n$ in $L^\infty(D)$ is the essential supremum of ${|g_n|,}$ as in Figure 6:

$$\begin{aligned} \|g_n\|_{L^\infty(D)} &= \operatorname*{ess\,sup}_D |g_n| \\ &= \inf \{ M \in \mathbb{R} : \lambda(\{ |g_n| > M \}) = 0 \}, \end{aligned}$$

hence ${\|g_n\|_{L^\infty(D)} \ge 1 - \varepsilon}$ for every ${\varepsilon > 0.}$

n

3

ε

0.30

Figure 6. The drawing of Remark 13: $g_n$ on ${[0, 2],}$ dashed on the left of ${0,}$ three levels $M$ at or above 1 and the level ${M = 1 - \varepsilon.}$ The part of the graph above this level is thick, and the set ${\{ |g_n| > 1 - \varepsilon \}}$ near 2 is hatched on the axis. The sliders set $n$ and ${\varepsilon.}$

A sequence with $\sqrt{n}$#

Example 14. On ${(-1, \infty)}$ with the Lebesgue measure ${\lambda,}$ for every integer ${n \ge 1}$ let ${f_n \colon (-1, \infty) \to \mathbb{R}}$ be given by

$$f_n(x) = \frac{\sqrt{n + 1}}{\sqrt{n} \, (x^2 + 2x) + \sqrt{n + \sqrt{n}} \, (2x + 3)}.$$

In the denominator of the functions $f_n$ of Example 14, ${x^2 + 2x}$ is a parabola with minimum at ${x = -1,}$ hence ${x^2 + 2x \ge 1 - 2 = -1.}$ Moreover, ${2x + 3 > -2 + 3 = 1}$ for every ${x \in (-1, \infty).}$ Hence, since ${\sqrt{n} \ge 0,}$ the denominator of $f_n$ satisfies

$$\sqrt{n} \, (x^2 + 2x) + \sqrt{n + \sqrt{n}} \, (2x + 3) \ge -\sqrt{n} + \sqrt{n + \sqrt{n}} \ge -\sqrt{n} + \sqrt{n} = 0,$$

and ${f_n \ge 0}$ on ${(-1, +\infty).}$ Indeed, the first inequality is strict, because ${2x + 3 > 1,}$ so the denominator is positive.

(1) Study the pointwise convergence of $(f_n)$ on ${(-1, \infty).}$ For every ${x \in (-1, \infty),}$ factoring out $\sqrt{n}$ in the denominator,

$$\begin{aligned} f_n(x) &= \frac{\sqrt{n + 1}}{\sqrt{n} \, (x^2 + 2x) + \sqrt{n + \sqrt{n}} \, (2x + 3)} \\ &= \frac{\sqrt{n + 1}}{\sqrt{n} \Big( x^2 + 2x + \frac{\sqrt{n + \sqrt{n}}}{\sqrt{n}} \, (2x + 3) \Big)}. \end{aligned}$$

Now ${\sqrt{n + 1}/\sqrt{n} \to 1}$ and ${\sqrt{n + \sqrt{n}}/\sqrt{n} = \sqrt{1 + \sqrt{n}/n} \to 1,}$ hence, for every ${x \in (-1, \infty),}$

$$\begin{aligned} f_n(x) \xrightarrow[n \to \infty]{} \frac{1}{x^2 + 2x + 2x + 3} &= \frac{1}{x^2 + 4x + 3} \\ &= \frac{1}{(x + 1)(x + 3)} =: f(x). \end{aligned}$$

Thus ${f_n \to f}$ pointwise on ${(-1, \infty).}$

(2) Study the convergence of $(f_n)$ in ${L^1((-1, \infty)).}$ By the uniqueness of limits, Remark 12 of the previous post, the candidate limit is ${f.}$ Notice that ${f(x) \sim 1/(2(x + 1))}$ as ${x \to -1^+,}$ and ${1/(2(x + 1)) \notin L^1((-1, 0)).}$ Hence ${f \notin L^1((-1, \infty)).}$ By Remark 12, it is impossible that ${f_n \to f}$ in ${L^1((-1, \infty)),}$ because that would imply ${f \in L^1((-1, \infty)).}$ Indeed, Remark 12 applies: the denominator of $f_n$ is at least ${\sqrt{n + \sqrt{n}} - \sqrt{n} > 0}$ on ${(-1, \infty)}$ and at least ${\sqrt{n} \, x^2}$ for ${x > 0,}$ so each $f_n$ belongs to ${L^1((-1, \infty)).}$

(3) Study the convergence of $(f_n)$ in ${L^1((1, \infty)).}$ For every ${x \in (1, \infty),}$

$$\begin{aligned} |f_n(x)| = f_n(x) &= \frac{\sqrt{n + 1}}{\sqrt{n} \, (x^2 + 2x) + \sqrt{n + \sqrt{n}} \, (2x + 3)} \\ &= \frac{\sqrt{n + 1}}{\sqrt{n}} \, \frac{1}{x^2 + 2x + \frac{\sqrt{n + \sqrt{n}}}{\sqrt{n}} \, (2x + 3)} \\ &\le \sqrt{2} \, \frac{1}{x^2 + 4x + 3} \\ &= \sqrt{2} \, \frac{1}{(x + 1)(x + 3)}, \end{aligned}$$

because the first factor and the coefficient of ${2x + 3}$ satisfy, using ${1 \le n}$ for the first one,

$$\frac{\sqrt{n + 1}}{\sqrt{n}} \le \frac{\sqrt{n + n}}{\sqrt{n}} = \frac{\sqrt{2} \, \sqrt{n}}{\sqrt{n}} = \sqrt{2}, \qquad \frac{\sqrt{n + \sqrt{n}}}{\sqrt{n}} \ge \frac{\sqrt{n}}{\sqrt{n}} = 1.$$

The function ${\sqrt{2}/((x + 1)(x + 3))}$ belongs to ${L^1((1, \infty)).}$ Moreover, ${f_n \to f}$ pointwise by (1). Hence the dominated convergence theorem, Theorem 8 of the previous post with ${p = 1,}$ gives ${f_n \to f}$ in ${L^1((1, \infty))}$ as ${n \to \infty.}$