Theoretical Recall
Let
\[A \subset \mathbb{R}, \qquad fₙ, \qquad fₙ : A \to \mathbb{R}, \qquad f : A \to \mathbb{R}.\]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:
\[\forall \varepsilon > 0, \ \exists n_{\varepsilon,x} : \forall n > n_{\varepsilon,x}, \ |f_n(x)-f(x)| < \varepsilon .\]Uniform convergence
The sequence fₙ converges uniformly to f on A if
\[\forall \varepsilon > 0,\ \exists n_\varepsilon :\ \forall n > n_\varepsilon,\ \forall x \in A,\quad |f_n(x)-f(x)| < \varepsilon .\]Equivalently, setting
\[\alpha_n = \sup_{x \in A} |f_n(x) - f(x)| ,\]we have
\[\alpha_n \to 0 .\]Facts
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Uniform convergence ⇒ pointwise convergence (not conversely).
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If fₙ ⇉ f and each fₙ is continuous on an interval I, then f is continuous on I.
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Limits commute with integrals:
- If moreover
then
\[f \in C^1(I), \qquad f' = g .\]Exercises
Note. When summing a power series, unless otherwise specified, sums are understood on intervals
\([x₀ - ρ,\, x₀ + ρ], \quad 0 \leq ρ \leq r\)
where r is the radius of convergence.
Ex 1. For x ∈ [0,1] and p ∈ ℝ, consider
\[f_n(x) = n^{p}\,x\,e^{-n x^{2}} .\]Study pointwise and uniform convergence (as p varies).
Solution:
By comparison of infinitesimals,
so fₙ → 0 pointwise.
For uniform convergence (the functions are continuous), compute the maximum:
\[f_n'(x) = n^{p} e^{-n x^{2}} (1 - 2n x^{2}).\]The maximum occurs at
\[x = \frac{1}{\sqrt{2n}} ,\]hence
\[\alpha_n = \sup_{x \in [0,1]} |f_n(x)| = \frac{e^{-1/2}}{\sqrt{2}} \, n^{\,p-\tfrac12}.\]Therefore
\[\lim_{n \to \infty} \alpha_n = \begin{cases} 0 & p < \tfrac12, \\[6pt] \dfrac{e^{-1/2}}{\sqrt{2}} & p = \tfrac12, \\[6pt] +\infty & p > \tfrac12 . \end{cases}\]Observation: The exponential decay dominates only if p < ½.
Final Result:
\[\text{Pointwise: } f_n \to 0. \quad \text{Uniform on } [0,1] \ \text{iff } p < \tfrac12 .\]Ex 2. For x ≥ 0,
\[f_n(x) = \sqrt[n]{\,n + x^{n}\,} .\]Study pointwise and uniform convergence.
Solution:
Pointwise limit:
For uniform convergence on [0,∞):
- On [0,1]:
- On [1,∞):
So gₙ is decreasing and
\[\alpha_{n,2} = \sup_{x\in[1,\infty)} |f_n(x)-x| = \sqrt[n]{n+1}-1 \to 0 .\]Thus
\[\alpha_n = \sup_{x\in[0,\infty)} |f_n(x)-f(x)| \to 0 .\]Observation: Uniform convergence holds globally, even though the pointwise limit function changes definition at x=1.
Final Result:
\[\text{Pointwise: } f_n \to f,\ f(x) = \begin{cases} 1 & x \in [0,1], \\ x & x > 1. \end{cases} \quad \text{Uniform on } [0,\infty) .\]Ex 3. For x ∈ [0,1],
\[f_n(x) = n^{2} x^{n} (1-x^{4}) .\]Solution:
Clearly
For uniform convergence,
\[f_n'(x) = n^{2} x^{n-1} \bigl( n - (n+4)x^{4} \bigr).\]The maximum is at
\[x = \left(\frac{n}{n+4}\right)^{1/4},\]hence
\[\alpha_n = n^{2} \left( \frac{n}{n+4} \right)^{n/4} \frac{4}{n+4} \;\to\; +\infty .\]Observation: The growth of n² prevents uniform convergence despite pointwise vanishing.
Final Result:
\[\text{Pointwise: } f_n \to 0, \qquad \text{Not uniform on } [0,1].\]Ex 4. For x ∈ [0,1] and p ∈ ℝ,
\[f_n(x) = n^{p} x^{n} (1-x^{2}) .\]Solution:
Pointwise,
Derivative:
\[f_n'(x) = n^{p} x^{n-1} \bigl( n - (n+2)x^{2} \bigr).\]Maximum at
\[x = \sqrt{\tfrac{n}{n+2}} .\]So
\[\alpha_n = n^{p} \left(\frac{n}{n+2}\right)^{n/2} \frac{2}{n+2}.\]Hence
\[\alpha_n \to \begin{cases} 0 & p < 1, \\[6pt] \dfrac{2}{e} & p = 1, \\[6pt] +\infty & p > 1 . \end{cases}\]Observation: The balance between polynomial growth and exponential decay breaks precisely at p=1.
Final Result:
\[\text{Pointwise: } f_n \to 0, \qquad \text{Uniform on } [0,1] \ \text{iff } p < 1 .\]Ex 5. For x ∈ [0,1],
\[f_n(x) = \frac{\sin(nx)}{n} .\]Solution:
By notable limits,
so pointwise convergence holds.
Since
\[|f_n(x)| \le \frac{1}{n},\]we have
\[\alpha_n \le \frac{1}{n} \to 0 .\]Observation: The sine oscillation, bounded by 1, ensures uniform convergence at the same rate as 1/n.
Final Result:
\[\text{Pointwise: } f_n \to 0, \qquad \text{Uniform on } [0,1].\]Ex 6. For x ∈ ℝ,
\[f_n(x) = \arctan(nx) .\]Solution:
Pointwise limit:
No uniform convergence on ℝ since the limit is discontinuous.
For any a > 0:
\[\sup_{x \le -a} \left| f_n(x) + \frac{\pi}{2} \right| = \frac{\pi}{2} - \arctan(na) \to 0 ,\] \[\sup_{x \ge a} \left| f_n(x) - \frac{\pi}{2} \right| = \frac{\pi}{2} - \arctan(na) \to 0 .\]Observation: Uniform convergence holds only away from the discontinuity at x=0.
Final Result:
\[\text{No uniform convergence on } \mathbb{R}, \quad \text{Uniform on } [a,\infty),\ (-\infty,-a] \ \forall a>0 .\]Ex 7. Study convergence and compute the sum of
\[\sum_{n=1}^{\infty}\frac{n+1}{n!}\,x^{n},\qquad x\in\mathbb{R}.\]Solution:
Ratio test:
so absolute convergence for all x.
Split:
\[\sum_{n=1}^\infty \frac{n}{n!} x^{n} = x \sum_{n=1}^\infty \frac{x^{n-1}}{(n-1)!} = x e^{x},\] \[\sum_{n=1}^\infty \frac{x^{n}}{n!} = e^{x} - 1 .\]Thus
\[s(x) = x e^{x} + e^{x} - 1 .\]Observation: The manipulation uses termwise differentiation/integration of power series.
Final Result:
\[\sum_{n=1}^{\infty}\frac{n+1}{n!}\,x^{n} = e^{x}(1+x) - 1 .\]Ex 8. Study the series
\[\sum_{n=1}^{\infty}(2^{n}+3^{n})\,x^{n}.\]Solution:
Ratio test:
At x=⅓:
\[\sum \frac{2^{n}+3^{n}}{3^{n}}\]diverges.
At x=−⅓:
\[\sum \frac{2^{n}+3^{n}}{3^{n}}(-1)^{n}\]does not converge.
Observation: Even if the radius is finite, both endpoints fail here.
Final Result:
\[\text{Convergence set } X = \Bigl(-\tfrac{1}{3},\tfrac{1}{3}\Bigr).\]Ex 9. Study and sum
\[\sum_{n=1}^{\infty}(-1)^{n}\,n\,x^{2n-1}.\]Solution:
Root test:
| For | x | <1: |
Observation: Differentiating a simple geometric series is the key step.
Final Result:
\[\sum_{n=1}^{\infty}(-1)^{n}n x^{2n-1} = -\frac{x}{(1+x^{2})^{2}},\qquad |x|<1 .\]Ex 10. Study and sum
\[\sum_{n=0}^{\infty}\frac{x^{n}}{(n+1)(n+2)}.\]Solution:
Root test:
Use double integration:
\[\sum_{n=0}^{\infty}\frac{x^{n}}{(n+1)(n+2)} = \frac{1}{x^{2}}\int_{0}^{x}\int_{0}^{y}\sum_{n=0}^{\infty}t^{n}\,dt\,dy = \frac{1}{x^{2}}\int_{0}^{x}\int_{0}^{y}\frac{1}{1-t}\,dt\,dy = \frac{(1-x)\log(1-x)}{x^{2}} + \frac{1}{x}.\]Observation: Double integration allows to reduce the denominator (n+1)(n+2).
Final Result:
\[\sum_{n=0}^{\infty}\frac{x^{n}}{(n+1)(n+2)} = \frac{(1-x)\,\log(1-x)}{x^{2}} + \frac{1}{x}.\]