Solved Exercises — Sequences and Series of Functions
Theoretical Recall
Let:
\[A\subset\mathbb{R},\]and let:
\[f_n:A\to\mathbb{R},\] \[f:A\to\mathbb{R}.\]Pointwise Convergence
The sequence fₙ converges pointwise to f on A if:
\[\lim_{n\to\infty}f_n(x)=f(x), \qquad \forall x\in A.\]Equivalently:
\[\forall\varepsilon>0, \quad \exists n_{\varepsilon,x}: \quad \forall n>n_{\varepsilon,x}, \quad |f_n(x)-f(x)|<\varepsilon.\]The index n depends both on ε and on the chosen point x.
Uniform Convergence
The sequence fₙ converges uniformly to f on A if:
\[\forall\varepsilon>0, \quad \exists n_\varepsilon: \quad \forall n>n_\varepsilon, \quad \forall x\in A, \quad |f_n(x)-f(x)|<\varepsilon.\]Unlike pointwise convergence, the same index n works simultaneously for every x in A.
Equivalently, setting:
\[\alpha_n = \sup_{x\in A}|f_n(x)-f(x)|,\]uniform convergence is equivalent to:
\[\alpha_n\to0.\]Fundamental Facts
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Uniform convergence implies pointwise convergence, but the converse does not hold.
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If fₙ converges uniformly to f and every fₙ is continuous on an interval I, then f is continuous on I.
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On a compact interval [a,b], uniform convergence allows the limit to commute with integration:
- Suppose:
and:
\[f_n\in C^1(I).\]Under the standard hypotheses ensuring convergence of fₙ at at least one point of I, it follows that:
\[f\in C^1(I),\]and:
\[f'=g.\]Worked Exercises
Note. When summing a power series, unless otherwise specified, sums are understood on intervals:
\[[x_0-\rho,x_0+\rho], \qquad 0\le\rho\le r,\]where r is the radius of convergence.
Exercise 1
For x ∈ [0,1] and p ∈ ℝ, consider:
\[f_n(x)=n^p x e^{-nx^2}.\]Study pointwise and uniform convergence as p varies.
Solution.
For x = 0:
\[f_n(0)=0.\]For every fixed x > 0, the exponential factor dominates every power of n:
\[n^p x e^{-nx^2}\to0.\]Therefore:
\[\lim_{n\to\infty}f_n(x)=0\]for every x ∈ [0,1].
Thus fₙ converges pointwise to zero.
For uniform convergence, compute the derivative:
\[f_n'(x) = n^p e^{-nx^2}(1-2nx^2).\]The interior critical point is determined by:
\[1-2nx^2=0.\]Hence:
\[x=\frac{1}{\sqrt{2n}}.\]At this point:
\[\alpha_n = \sup_{x\in[0,1]}|f_n(x)|.\]Therefore:
\[\alpha_n = n^p \frac{1}{\sqrt{2n}} e^{-1/2}.\]Thus:
\[\alpha_n = \frac{e^{-1/2}}{\sqrt2} n^{p-\frac12}.\]Consequently:
\[\lim_{n\to\infty}\alpha_n = \begin{cases} 0, & p<\frac12,\\ \dfrac{e^{-1/2}}{\sqrt2}, & p=\frac12,\\ +\infty, & p>\frac12. \end{cases}\]Observation: The exponential decay dominates only if p < 1/2.
Final Result
\[f_n\to0 \text{ pointwise on }[0,1], \qquad f_n\to0 \text{ uniformly iff }p<\frac12.\]Exercise 2
For x ≥ 0:
\[f_n(x)=\sqrt[n]{n+x^n}.\]Study pointwise and uniform convergence.
Solution.
For 0 ≤ x ≤ 1:
\[n \le n+x^n \le n+1.\]Taking n-th roots:
\[n^{1/n} \le f_n(x) \le (n+1)^{1/n}.\]Both bounds tend to 1, so:
\[f_n(x)\to1.\]For x > 1:
\[f_n(x) = x\left(1+\frac{n}{x^n}\right)^{1/n}.\]Since:
\[\frac{n}{x^n}\to0,\]we obtain:
\[f_n(x)\to x.\]Thus the pointwise limit is:
\[f(x) = \begin{cases} 1, & 0\le x\le1,\\ x, & x>1. \end{cases}\]For uniform convergence on [0,1]:
\[\alpha_{n,1} = \sup_{x\in[0,1]}|f_n(x)-1|.\]Since fₙ is increasing in x:
\[\alpha_{n,1} = (n+1)^{1/n}-1.\]Hence:
\[\alpha_{n,1}\to0.\]On [1,+∞), define:
\[g_n(x) = (n+x^n)^{1/n}-x.\]Its derivative is:
\[g_n'(x) = x^{n-1}(n+x^n)^{\frac1n-1}-1.\]Equivalently:
\[g_n'(x) = \left(1+\frac{n}{x^n}\right)^{\frac{1-n}{n}}-1.\]For x ≥ 1:
\[g_n'(x)<0.\]Therefore gₙ is decreasing and its maximum occurs at x = 1:
\[\alpha_{n,2} = \sup_{x\in[1,\infty)}|f_n(x)-x|.\]Hence:
\[\alpha_{n,2} = (n+1)^{1/n}-1.\]Therefore:
\[\alpha_{n,2}\to0.\]Combining the two regions:
\[\alpha_n = \sup_{x\in[0,\infty)}|f_n(x)-f(x)| \to0.\]Observation: Uniform convergence holds globally, even though the pointwise limit function changes definition at x = 1.
Final Result
\[f_n\to f \text{ uniformly on }[0,\infty), \qquad f(x)= \begin{cases} 1, & 0\le x\le1,\\ x, & x>1. \end{cases}\]Exercise 3
For x ∈ [0,1]:
\[f_n(x) = n^2x^n(1-x^4).\]Solution.
For every fixed x ∈ [0,1):
\[x^n\to0.\]At x = 1:
\[f_n(1)=0.\]Therefore:
\[f_n(x)\to0\]pointwise on [0,1].
Differentiate:
\[f_n'(x) = n^2x^{n-1} \left(n-(n+4)x^4\right).\]The maximum occurs when:
\[x^4 = \frac{n}{n+4}.\]Thus:
\[x = \left(\frac{n}{n+4}\right)^{1/4}.\]At this point:
\[\alpha_n = n^2 \left(\frac{n}{n+4}\right)^{n/4} \frac{4}{n+4}.\]Now:
\[\left(\frac{n}{n+4}\right)^{n/4} = \left(1+\frac4n\right)^{-n/4} \to e^{-1}.\]Moreover:
\[\frac{4n^2}{n+4}\sim4n.\]Therefore:
\[\alpha_n\to+\infty.\]Observation: The growth of n² prevents uniform convergence despite pointwise vanishing.
Final Result
\[f_n\to0 \text{ pointwise on }[0,1], \qquad \text{the convergence is not uniform}.\]Exercise 4
For x ∈ [0,1] and p ∈ ℝ:
\[f_n(x) = n^p x^n(1-x^2).\]Solution.
For every x ∈ [0,1):
\[x^n\to0.\]At x = 1:
\[f_n(1)=0.\]Hence:
\[f_n(x)\to0\]pointwise.
Differentiate:
\[f_n'(x) = n^p x^{n-1} \left(n-(n+2)x^2\right).\]The maximum occurs at:
\[x = \sqrt{\frac{n}{n+2}}.\]Therefore:
\[\alpha_n = n^p \left(\frac{n}{n+2}\right)^{n/2} \frac{2}{n+2}.\]Since:
\[\left(\frac{n}{n+2}\right)^{n/2} = \left(1+\frac2n\right)^{-n/2} \to e^{-1},\]we have:
\[\alpha_n \sim \frac{2}{e}n^{p-1}.\]Hence:
\[\alpha_n\to \begin{cases} 0, & p<1,\\ \dfrac2e, & p=1,\\ +\infty, & p>1. \end{cases}\]Observation: The balance between polynomial growth and exponential decay breaks precisely at p = 1.
Final Result
\[f_n\to0 \text{ pointwise on }[0,1], \qquad f_n\to0 \text{ uniformly iff }p<1.\]Exercise 5
For x ∈ [0,1]:
\[f_n(x) = \frac{\sin(nx)}{n}.\]Solution.
For every x:
\[|\sin(nx)|\le1.\]Therefore:
\[|f_n(x)| \le \frac1n.\]Since:
\[\frac1n\to0,\]we obtain pointwise convergence:
\[f_n(x)\to0.\]Moreover:
\[\alpha_n = \sup_{x\in[0,1]}|f_n(x)| \le \frac1n.\]Therefore:
\[\alpha_n\to0.\]Observation: The sine oscillation, bounded by 1, ensures uniform convergence at the same rate as 1/n.
Final Result
\[f_n\to0 \text{ uniformly on }[0,1].\]Exercise 6
For x ∈ ℝ:
\[f_n(x)=\arctan(nx).\]Solution.
For x > 0:
\[nx\to+\infty,\]so:
\[\arctan(nx)\to\frac{\pi}{2}.\]For x = 0:
\[f_n(0)=0.\]For x < 0:
\[nx\to-\infty,\]so:
\[\arctan(nx)\to-\frac{\pi}{2}.\]Therefore:
\[f(x) = \begin{cases} \dfrac{\pi}{2}, & x>0,\\ 0, & x=0,\\ -\dfrac{\pi}{2}, & x<0. \end{cases}\]Every fₙ is continuous on ℝ, while the pointwise limit f is discontinuous at x = 0.
Therefore the convergence cannot be uniform on ℝ.
For any a > 0:
\[\sup_{x\ge a} \left| f_n(x)-\frac{\pi}{2} \right| = \frac{\pi}{2}-\arctan(na).\]Hence:
\[\frac{\pi}{2}-\arctan(na)\to0.\]Similarly:
\[\sup_{x\le-a} \left| f_n(x)+\frac{\pi}{2} \right| = \frac{\pi}{2}-\arctan(na) \to0.\]Observation: Uniform convergence holds only away from the discontinuity at x = 0.
Final Result
\[\text{The convergence is not uniform on }\mathbb{R}, \text{ but it is uniform on } [a,\infty) \text{ and } (-\infty,-a] \text{ for every }a>0.\]Exercise 7
Study convergence and compute the sum:
\[\sum_{n=1}^{\infty} \frac{n+1}{n!}x^n.\]Solution.
The factorial in the denominator implies an infinite radius of convergence:
\[r=+\infty.\]Therefore the series converges absolutely for every real x.
Split the series:
\[\sum_{n=1}^{\infty} \frac{n+1}{n!}x^n = \sum_{n=1}^{\infty} \frac{n}{n!}x^n + \sum_{n=1}^{\infty} \frac{x^n}{n!}.\]For the first term:
\[\frac{n}{n!} = \frac{1}{(n-1)!}.\]Thus:
\[\sum_{n=1}^{\infty} \frac{n}{n!}x^n = x \sum_{n=1}^{\infty} \frac{x^{n-1}}{(n-1)!}.\]Hence:
\[\sum_{n=1}^{\infty} \frac{n}{n!}x^n = xe^x.\]For the second term:
\[\sum_{n=1}^{\infty} \frac{x^n}{n!} = e^x-1.\]Therefore:
\[s(x) = xe^x+e^x-1.\]Observation: The manipulation uses the standard exponential power series and the algebraic identity n/n! = 1/(n−1)!.
Final Result
\[\sum_{n=1}^{\infty} \frac{n+1}{n!}x^n = e^x(1+x)-1\]Exercise 8
Study the series:
\[\sum_{n=1}^{\infty} (2^n+3^n)x^n.\]Solution.
Let:
\[a_n=2^n+3^n.\]Then:
\[\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = 3.\]Therefore the radius of convergence is:
\[r=\frac13.\]For:
\[|x|<\frac13,\]the series converges absolutely.
At:
\[x=\frac13,\]we obtain:
\[\sum_{n=1}^{\infty} \left[ \left(\frac23\right)^n+1 \right].\]The general term does not tend to zero, so the series diverges.
At:
\[x=-\frac13,\]we obtain:
\[\sum_{n=1}^{\infty} (-1)^n \left[ \left(\frac23\right)^n+1 \right].\]Again, the general term does not tend to zero.
Therefore the series diverges.
Observation: Even if the radius is finite, both endpoints fail here.
Final Result
\[X= \left( -\frac13, \frac13 \right)\]Exercise 9
Study and sum:
\[\sum_{n=1}^{\infty} (-1)^n n x^{2n-1}.\]Solution.
The convergence condition is:
\[|x|<1.\]At x = 1 and x = −1, the general term does not tend to zero, so both endpoints are excluded.
Hence:
\[X=(-1,1).\]| For | x | < 1, start from the geometric series: |
Differentiate term by term:
\[\sum_{n=1}^{\infty} 2n(-1)^n x^{2n-1} = -\frac{2x}{(1+x^2)^2}.\]Dividing by 2:
\[\sum_{n=1}^{\infty} n(-1)^n x^{2n-1} = -\frac{x}{(1+x^2)^2}.\]Observation: Differentiating a simple geometric series is the key step.
Final Result
\[\sum_{n=1}^{\infty} (-1)^n n x^{2n-1} = -\frac{x}{(1+x^2)^2}, \qquad |x|<1\]Exercise 10
Study and sum:
\[\sum_{n=0}^{\infty} \frac{x^n}{(n+1)(n+2)}.\]Solution.
The radius of convergence is:
\[r=1.\]At x = 1:
\[\sum_{n=0}^{\infty} \frac{1}{(n+1)(n+2)}\]converges.
Indeed:
\[\frac{1}{(n+1)(n+2)} = \frac{1}{n+1} - \frac{1}{n+2}.\]Thus the series is telescopic.
At x = −1:
\[\sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1)(n+2)}\]converges absolutely because:
\[\sum_{n=0}^{\infty} \frac{1}{(n+1)(n+2)}\]converges.
Therefore:
\[X=[-1,1].\]| For | x | < 1, begin with: |
Integrating from 0 to y:
\[\sum_{n=0}^{\infty} \frac{y^{n+1}}{n+1} = -\log(1-y).\]Integrating again from 0 to x:
\[\sum_{n=0}^{\infty} \frac{x^{n+2}}{(n+1)(n+2)} = \int_0^x-\log(1-y)\,dy.\]The integral is:
\[\int_0^x-\log(1-y)\,dy = x+(1-x)\log(1-x).\]Therefore, for x ≠ 0:
\[\sum_{n=0}^{\infty} \frac{x^n}{(n+1)(n+2)} = \frac{x+(1-x)\log(1-x)}{x^2}.\]Equivalently:
\[\sum_{n=0}^{\infty} \frac{x^n}{(n+1)(n+2)} = \frac{1}{x} + \frac{(1-x)\log(1-x)}{x^2}.\]At x = 0, the original series gives:
\[\frac{1}{(0+1)(0+2)} = \frac12.\]The expression above has a removable singularity at x = 0, and its continuous extension has value 1/2.
At x = 1:
\[\sum_{n=0}^{\infty} \frac{1}{(n+1)(n+2)} = 1.\]At x = −1, using the continuous endpoint value of the sum:
\[\sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1)(n+2)} = 2\log2-1.\]Observation: Double integration allows us to produce the denominator (n+1)(n+2) from the geometric series.
Final Result
\[\sum_{n=0}^{\infty} \frac{x^n}{(n+1)(n+2)} = \begin{cases} \dfrac{x+(1-x)\log(1-x)}{x^2}, & -1\le x<1,\ x\ne0,\\[8pt] \dfrac12, & x=0,\\[8pt] 1, & x=1. \end{cases}\]Continue Exploring Calculus
Sequences and series of functions extend the idea of convergence from numbers to entire functions. The distinction between pointwise and uniform convergence becomes essential when studying continuity, integration, differentiation, and power series.
Power series also connect infinite processes with familiar elementary functions, allowing complicated sums to be studied through algebra, differentiation, and integration.