Homework 2

1. (20 points) Show that the following series of functions converge uniformly on \([1, +\infty)\).

   1. \(\sum_{n=1}^{\infty} \frac{1}{(x+1)^n}\)

      Solution

      Let's use M-test

      Consider \(\left|\frac{1}{(x+1)^{n}}\right|\), we know \(x\in [1,+\infty)\), then \(x+1\in [2,\infty)\), then \((x+1)^{n}\in[2^{n},+\infty)\)​

      Then we have \(\left|\frac{1}{(x+1)^{n}}\right|\leq\frac{1}{2^{n}}=M_{n}\)​

      And we know \(\sum M_n\) is geometric series and it's convergent

      Thus by M-test we have \(\sum_{n=1}^{\infty} \frac{1}{(x+1)^n}\) is convergent uniformly 2. \(\sum_{n=1}^{\infty}\frac{1}{n(n+x)}\)

      Solution

      Consider \(\left|\frac{1}{n(n+x)}\right|\), we know \(x\in [1,+\infty)\), then \(x+n\in[n+1,\infty)\), then \(n\left(n+x)\in[n^{2}+n,+\infty\right)\)

      Then we have \(\left|\frac{1}{n(n+x)}\right|\leq\frac{1}{n^{2}+n}\leq\frac{1}{n^{2}}\leq\frac{1}{n\left(n-1\right)} =M_{n}\)

      And we know \(\sum M_n\) is telescope sum and it's convergent

      Thus by M-test we have \(\sum_{n=1}^{\infty}\frac{1}{n(n+x)}\) is convergent uniformly

2. (20 points) Prove that, if the series of functions \(\sum_{n=1}^{\infty} f_n\) converges uniformly, then the sequence \((f_n)\) converges uniformly to the zero function.

   Proof

   We are going to prove \(\forall \varepsilon>0,\exists N\in \N,\forall n\geq N:|f_{n}(x)-0|<\varepsilon,\forall x\)​

   Thus consider \(|f_n(x)|\), then we need to bound \(f_n(x)\)​

   We know \(\sum_{n=1}^{\infty} f_n\) converges uniformly, then use Cauchy Criterium: \(\forall\varepsilon^{\prime}>0\), \(\exists N^{\prime}\in\mathbb{N},\forall n>N^{\prime}\geq N^{\prime}:\) \(|f_{n}(x)+f_{N^{\prime}}(x)|<\varepsilon^{\prime}\), \(\forall x\)

   Let \(\varepsilon^{\prime}=\varepsilon-\left|f_{N'}\left(x\right)\right|\). Then \(|f_{n}(x)|<\varepsilon^{\prime}+|f_{N^{\prime}}(x)|=\varepsilon,\forall x\)

3. (30 points) Let \(f(x) = \sum_{n=1}^{\infty}\frac{1}{n}\sin \left(\frac{x}{n+1}\right)\). Show that \(f(x)\) converges uniformly on any closed, bounded interval \([a, b]\) and it satisfies \(|f(x)| \leq |x| \quad \text{and}\quad |f'(x)| \leq 1.\)

   Proof

   Use Cauchy Criterium:

   Consider \(|\frac{1}{n}\sin\left(\frac{x}{n+1}\right)+\frac{1}{n+1}\sin\left(\frac{x}{n+2}\right )+\dots+\frac{1}{m}\sin\left(\frac{x}{m+1}\right)|<\frac{1}{n}|\sin\left(\frac{x}{n+1} \right)+\sin\left(\frac{x}{n+2}\right)+\dots+\sin\left(\frac{x}{m+1}\right)|\)​

   \(\leq\frac{1}{n}\left(\left|\sin\frac{x}{n+1}\right|+\left|\sin\frac{x}{n+2}\right |+\cdots+\left|\sin\frac{x}{m+1}\right|\right)\leq\frac{1}{n}\left(\left|\frac{x}{n+1} \right|+\cdots+\left|\frac{x}{m+1}\right|\right)\leq\frac{1}{n}\left(n-m+1\right) \left|\frac{x}{n+1}\right|\)

   \(=\frac{n-m+1}{n^{2}+n}\left|x\right|\leq\frac{n+1}{n}\left|x\right|\leq\left|x\right |+\frac{\left|x\right|}{N}\)

   Since we know \(x\) in a bounded and closed set, thus \(|x|\leq M\), thus \(\left|x\right|+\frac{\left|x\right|}{N}\leq M\left(1+\frac{1}{N}\right)\)​

   Thus we can take \(N>\frac{M}{\varepsilon-M}\), then we get \(\forall\varepsilon > 0\), \(\exists N>\frac{M}{\varepsilon-M}\in\mathbb{N},\forall n>m\geq N\) such that \(|\frac{1}{n}\sin\left(\frac{x}{n+1}\right)+\frac{1}{n+1}\sin\left(\frac{x}{n+2}\right )+\dots+\frac{1}{m}\sin\left(\frac{x}{m+1}\right)|<M\left(1+\frac{1}{N}\right)=\varepsilon\), \(\forall x \in A\).

   Thus \(f(x)\) converges uniformly

   ---

   Since we know \(\left|\frac{1}{n}\sin\left(\frac{x}{n+1}\right)\right|\leq\frac{1}{n}\left|\frac{x}{n+1} \right|\leq\frac{\left|x\right|}{n\left(n+1\right)}=\frac{\left|x\right|}{n}-\frac{\left|x\right|}{n+1}\), then \(f(x)=\sum_{n=1}^{\infty}\frac{1}{n}\sin\left(\frac{x}{n+1}\right)=\frac{\left|x\right|}{1} -\frac{\left|x\right|}{1+1}+\frac{\left|x\right|}{2}-\frac{\left|x\right|}{2+1}+\cdots +\frac{\left|x\right|}{n}-\frac{\left|x\right|}{n+1}+\cdots=\frac{\left|x\right|}{1} +\lim_{n\to\infty}\left(\frac{\left|x\right|}{n}-\frac{\left|x\right|}{n+1}\right )=\left|x\right|\)

   ---

   Choose \(x_0=b\), then \(f(x_{0})=\sum_{n=1}^{\infty}\frac{1}{n}\sin\left(\frac{x_{0}}{n+1}\right)=\sum_{n=1} ^{\infty}\frac{1}{n}\sin\left(\frac{b}{n+1}\right)\)​

   We know that \(\left|\frac{1}{n}\sin(\frac{b}{n+1})\right|\leq\frac{\left|b\right|}{n\left(n+1\right)}\) and \(\sum\frac{\left|b\right|}{n\left(n+1\right)}\) since p-series

   Thus by comparison test, \(f(x_0)\) is convergent

   Also we prove \(\sum_{n=1}^{\infty}\left\lbrack\frac{1}{n}\sin\left(\frac{x}{n+1}\right)\right\rbrack ^{\prime}=\sum_{n=1}^{\infty}\frac{1}{n}\cos\left(\frac{x}{n+1}\right)\cdot\left( \frac{1}{n+1}\right)=\sum_{n=1}^{\infty}\frac{1}{n\left(n+1\right)}\cos\left(\frac{x}{n+1} \right)\) is uniformly convergent

   Consider \(\left|\frac{1}{n\left(n+1\right)}\cos\left(\frac{x}{n+1}\right)\right|\leq\frac{1}{n\left(n+1\right)} =M_{n}\) and we know \(\sum M_n\) is convergent since p-series

   Thus \(\sum_{n=1}^{\infty}\left\lbrack\frac{1}{n}\sin\left(\frac{x}{n+1}\right)\right\rbrack ^{\prime}\) is uniformly convergent

   ---

   Thus we can use differentiability term by term, \(f^{\prime}(x)=\sum_{n=1}^{\infty}\frac{1}{n\left(n+1\right)}\cos\left(\frac{x}{n+1} \right)\)​

   Then \(\left|f^{\prime}(x)\right|=\sum_{n=1}^{\infty}\left|\frac{1}{n\left(n+1\right)}\cos \left(\frac{x}{n+1}\right)\right|\leq\sum\frac{1}{n\left(n+1\right)}=\frac{1}{1}- \frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdots+\frac{1}{n}-\frac{1}{n+1}+\cdots=1+\lim _{n\to\infty}\frac{1}{n\left(n+1\right)}=1\)​

4. (30 points) Find the domain of convergence for the following power series:

   1. \(\sum_{n=1}^{\infty}\frac{\ln(n+1)}{n+1}(x-2)^{n}\)

      Let's use D'A Criterium

      Consider \(\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right|=\lim_{n\to\infty}\left|\frac{\frac{\ln(n+2)}{n+2}(x-2)^{n+1}}{\frac{\ln(n+1)}{n+1}(x-2)^{n}} \right|=\lim_{n\to\infty}\left|\frac{\frac{\ln(n+2)}{n+2}(x-2)}{\frac{\ln(n+1)}{n+1}} \right|=\left|x-2\right|\lim_{n\to\infty}\frac{\ln(n+2)\cdot(n+1)}{\ln(n+1)\cdot(n+2)} =\left|x-2\right|<1\)​

      Then \(x\in (1,3)\)

      ---

      When \(x=1\), \(\sum_{n=1}^{\infty}\frac{\ln(n+1)}{n+1}(x-2)^{n}=\sum_{n=1}^{\infty}\frac{\ln(n+1)}{n+1} (-1)^{n}\)

      By alternating test, \(\lim_{n\to\infty}\frac{\ln(n+1)}{n+1}=\lim_{n\to\infty}\frac{\frac{1}{n+1}}{1}=0\) and \(\left\lbrack\frac{\ln(n+1)}{n+1}\right\rbrack^{\prime}=\frac{\frac{1}{n+1}\cdot\left(n+1\right)-\ln(n+1)}{\left(n+1\right)^{2}} =\frac{1-\ln(n+1)}{\left(n+1\right)^{2}}\)

      And we know \(n=1\Rightarrow\ln(n+1)=\ln(2)<1,\quad n\geq2\Rightarrow\ln(n+1)\geq\ln(3)\geq1\)

      Thus \(\left\lbrack\frac{\ln(n+1)}{n+1}\right\rbrack^{\prime}>0\) when \(n\geq 2\), which means \(\frac{\ln(n+1)}{n+1}\) is decreasing when \(n\geq 2\)

      And when \(n=1,\frac{\ln(n+1)}{n+1}=\frac{\ln2}{2}\). When \(n=2,\frac{\ln(n+1)}{n+1}=\frac{\ln3}{3}\). Here \(\frac{\ln2}{2}<\frac{\ln3}{3}\)

      Thus when \(x=1\), the series is convergent

      ---

      When \(x=3\), \(\sum_{n=1}^{\infty}\frac{\ln(n+1)}{n+1}(1)^{n}=\sum_{n=1}^{\infty}\frac{\ln(n+1)}{n+1}\) and we know \(\frac{1}{n+1}\leq\frac{\ln(n+1)}{n+1}\) when \(n>2\), \(\sum \frac1n\) is divergent

      Thus \(\sum_{n=1}^{\infty}\frac{\ln(n+1)}{n+1}\) is divergent. Thus \([1,3]\) is the domain of convergence 2. \(\sum_{n=1}^{\infty}\frac{3^{n} + (-2)^{n}}{n}(x + 1)^{n}.\)

      Let's use D'A Criterium

      Consider \(\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right|=\lim_{n\to\infty}\left|\frac{\frac{3^{n+1}+(-2)^{n+1}}{n+1}(x+1)^{n+1}}{\frac{3^{n}+(-2)^{n}}{n}(x+1)^{n}} \right|=\lim_{n\to\infty}\left|\frac{\left(3^{n+1}+\left(-2\right)^{n+1}\right)\left(x+1\right)n}{\left(3^{n}+\left(-2\right)^{n}\right)\left(n+1\right)} \right|=\left|x+1\right|\lim_{n\to\infty}\frac{\left(3^{n+1}+\left(-2\right)^{n+1}\right)}{\left(3^{n}+\left(-2\right)^{n}\right)}\)

      \(=\left|x+1\right|\lim_{n\to\infty}\frac{1+\frac{\left(-2\right)^{n+1}}{3^{n+1}}}{\frac{1}{3}+\frac{(-2)^{n}}{3^{n+1}}} =|x+1|\cdot 3<1\)

      Then \(x\in\)​\(\left(-\frac{4}{3},-\frac{2}{3}\right)\)​

      When \(x=-\frac{4}{3}\), \(\sum_{n=1}^{\infty}\frac{3^{n}+(-2)^{n}}{n}(x+1)^{n}=\sum_{n=1}^{\infty}\frac{3^{n}+(-2)^{n}}{n} (-\frac{1}{3})^{n}=\sum_{n=1}^{\infty}\frac{3^{n}+(-2)^{n}}{n\left(-3\right)^{n}} =\sum_{n=1}^{\infty}\left(-1\right)^{n}\frac{1}{n}+\sum_{n=1}^{\infty}\frac{(-2)^{n}}{n\left(-3\right)^{n}}\)​

      Clearly, these two series is convergent, then \(x=-\frac{4}{3}\) is valid

      When \(x=-\frac{2}{3}\), \(\sum_{n=1}^{\infty}\frac{3^{n}+(-2)^{n}}{n}(x+1)^{n}=\sum_{n=1}^{\infty}\frac{3^{n}+(-2)^{n}}{n} (\frac{1}{3})^{n}=\sum_{n=1}^{\infty}\frac{3^{n}+(-2)^{n}}{n\left(3\right)^{n}}=\sum _{n=1}^{\infty}\frac{1}{n}+\sum_{n=1}^{\infty}\frac{(-2)^{n}}{n\left(3\right)^{n}}\)​

      But the first series is not convergent, thus \(x=-\frac{2}{3}\) is not valid

      Thus the domain of convergence is \(\left\lbrack-\frac{4}{3},-\frac{2}{3}\right)\)​

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