Sequences and series

Consider \[ \sum_{n=1}^\infty \frac{x^n}{\sqrt{n^2+1}}. \] If \(|x|<1\), then \[ 0\le \frac{|x|^n}{\sqrt{n^2+1}}\le |x|^n, \] so the series converges absolutely by comparison with a geometric series.

If \(|x|>1\), then \[ \frac{|x|^n}{\sqrt{n^2+1}} \sim \frac{|x|^n}{n}, \] which does not tend to \(0\); hence the series diverges.

At \(x=1\), \[ \frac1{\sqrt{n^2+1}}\ge \frac1{\sqrt2\,n}, \] so the series diverges by comparison with the harmonic series.

At \(x=-1\), the series is \[ \sum_{n=1}^\infty \frac{(-1)^n}{\sqrt{n^2+1}}. \] The positive terms \(1/\sqrt{n^2+1}\) decrease to \(0\), so the alternating-series test gives convergence.

Therefore the set of real \(x\) for which the series converges is \[ [-1,1). \]

Let \[ S_m=\sum_{i=0}^m(-1)^ia_i. \] Because \((a_i)\) is decreasing and positive, each difference \(a_{2j}-a_{2j+1}\) is nonnegative.

If \(m=2r+1\) is odd, then \[ S_m=(a_0-a_1)+(a_2-a_3)+\cdots +(a_{2r}-a_{2r+1})\ge0. \] If \(m=2r\) is even, then \[ S_m=(a_0-a_1)+\cdots +(a_{2r-2}-a_{2r-1})+a_{2r}\ge0. \] Thus \(S_m\ge0\) in all cases.

For the upper bound, if \(m=2r\), then \[ S_m=a_0-(a_1-a_2)-\cdots-(a_{2r-1}-a_{2r})\le a_0. \] If \(m=2r+1\), then \[ S_m=a_0-(a_1-a_2)-\cdots-(a_{2r-1}-a_{2r})-a_{2r+1}\le a_0. \] Therefore \[ 0\le S_m\le a_0 \] for every \(m\ge0\).

<1>1. Conditional convergence of \(\sum a_n\) can occur while \(\sum a_n^2\) converges.

<1>2. Conditional convergence of \(\sum a_n\) can also occur while \(\sum a_n^2\) diverges.

<1>3. If \(\sum a_n\) converges absolutely, then \(\sum a_n^2\) converges.

1. We want some power series centered at 4 with convergence radius 1, so something of the form \(\sum_n f(n)(x-4)^n\) with \(L = \lim \frac{a_{n+1}}{a_n} = 1\).

   Then, by plugging in \(5\) and \(3\), we find that we want \(\sum (-1)^n f(n)\) to converge but \(\sum f(n)\) to diverge. The canonical example of this is the harmonic series with \(f(n) = \frac 1 n\), so we can just take \[f(x) = \sum_n \frac{(x-4)^n}{n}. \qed\]

Since \[ e^{2x}=\sum_{n=0}^\infty \frac{(2x)^n}{n!}, \] we have \[ xe^{2x}=\sum_{n=0}^\infty \frac{2^n}{n!}x^{n+1}. \] Thus \[ T_5(x)=x+2x^2+2x^3+\frac43x^4+\frac23x^5. \]

Taylor’s theorem gives, for some \(\xi\) between \(0\) and \(x\), \[ f(x)-T_5(x)=\frac{f^{(6)}(\xi)}{6!}x^6. \] From the derivative formula, \[ f^{(6)}(t)=64te^{2t}+192e^{2t}=(64t+192)e^{2t}. \] On \(0\le t\le1/2\), this is increasing, so \[ |f^{(6)}(t)|\le224e. \] Also \(x^6\le(1/2)^6=1/64\). Therefore \[ |f(x)-T_5(x)|\le \frac{224e}{6!\,64}=\frac{7e}{1440} \] for every \(x\in[0,1/2]\).

The Taylor series about \(x=1\) is \[ \boxed{\ln x=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(x-1)^n}{n},\qquad |x-1|<1.} \] In particular, \[ \boxed{\ln(1.1)\approx 0.10} \] to two decimal places.

<1>1. Derive the series from the geometric series.

<1>2. Evaluate at \(x=1.1\) and control the error.

1. Denoting \(D^n f \definedas \dd{^n f}{x^n}\) and noting that \(D^1 D^n f = D^{n+1}f\), we have \[\begin{align*} D^0f &= xe^{2x} \\ D^1f &= e^{2x} + 2D^0f \\ D^2f &= 2e^{2x} + 2D^1f \\ D^3f &= 4e^{2x} + 2D^2f \\ \end{align*}\]

   and (claim) so we find that \[D^n f = 2^{n-1}e^{2x} + 2D^{n-1}f.\]

   This holds for \(n=1\): we have \(D^1 f = e^{2x} + 2xe^{2x} = 2^0e^{2x} + 2D^0 f\), and the inductive step holds exactly because \[ D^{n+1}f = DD^{n}f \\ = D(2^{n-1}e^{2x} + 2D^{n-1}f) \\ =2^n e^{2x} + 2D^n f. \] \(\qed\)

Goal: Find a closed formula for the \(n\)-th derivative \(f^{(n)}(x)\) of \(f(x) = x e^{2x}\) for all integers \(n \ge 0\), and prove it by induction.

<1>1. Claim: For all \(n \ge 0\), \(f^{(n)}(x) = (2^n x + n 2^{n-1}) e^{2x}\).

<1>2. Base Case (\(n = 0\)): \(f^{(0)}(x) = f(x) = (2^0 x + 0 \cdot 2^{-1}) e^{2x} = x e^{2x}\).

<1>3. Inductive Step: Assume the statement holds for some integer \(k \ge 0\), i.e., \(f^{(k)}(x) = (2^k x + k 2^{k-1}) e^{2x}\). Then it holds for \(k+1\). Proof: <2>1. Differentiating \(f^{(k)}(x)\) using the product rule: \[f^{(k+1)}(x) = \frac{d}{dx}\left[(2^k x + k 2^{k-1}) e^{2x}\right] = \left(\frac{d}{dx}(2^k x + k 2^{k-1})\right) e^{2x} + (2^k x + k 2^{k-1}) \left(\frac{d}{dx} e^{2x}\right).\] <2>2. Compute the components: \[\frac{d}{dx}(2^k x + k 2^{k-1}) = 2^k, \qquad \frac{d}{dx} e^{2x} = 2 e^{2x}.\] <2>3. Combine and factor out \(e^{2x}\): \[f^{(k+1)}(x) = 2^k e^{2x} + 2(2^k x + k 2^{k-1}) e^{2x} = \left(2 \cdot 2^k x + 2^k + 2 k 2^{k-1}\right) e^{2x}\] \[= \left(2^{k+1} x + 2^k(1 + k)\right) e^{2x} = \left(2^{k+1} x + (k+1) 2^{(k+1)-1}\right) e^{2x}.\] This matches the formula for \(n = k+1\).

<1>4. Alternative derivation using Leibniz’s Rule: Proof: <2>1. Leibniz’s Rule states that \((u v)^{(n)} = \sum_{k=0}^n \binom{n}{k} u^{(n-k)} v^{(k)}\). <2>2. Let \(u(x) = x\) and \(v(x) = e^{2x}\). Then \(u^{(0)}(x) = x\), \(u^{(1)}(x) = 1\), and \(u^{(j)}(x) = 0\) for all \(j \ge 2\). Also \(v^{(k)}(x) = 2^k e^{2x}\) for all \(k \ge 0\). <2>3. The sum has only two non-zero terms (\(k=n\) and \(k=n-1\) for \(u^{(0)}\) and \(u^{(1)}\)): \[f^{(n)}(x) = \binom{n}{0} x (2^n e^{2x}) + \binom{n}{1} (1) (2^{n-1} e^{2x}) = (2^n x + n 2^{n-1}) e^{2x}.\]

<1>5. Conclusion: \(f^{(n)}(x) = (2^n x + n 2^{n-1}) e^{2x} = 2^{n-1}(2x + n)e^{2x}\) for all \(n \ge 0\).

1. Suppose \(f_n \rightrightarrows g\) with each \(f_n\) bounded; we want to show that all of the \(f_n\) are uniformly bounded by some \(M\), i.e. \[ \exists M \suchthat \forall x\in \RR, \forall n\in \NN, \quad \abs{f_n(x)} \leq M. \]

• Since each \(f_n\) is bounded, we can produce some \(M_n\) such that \(\abs{f_n(x)} \leq M < \infty\).
• Since \(f_n \rightrightarrows g\), we can give ourselves an epsilon of room and get an \(N\) such that \(n\geq N \implies \abs{f_n(x) - g(x)} < \varepsilon\). We then write \[ f_n(x) = f_n(x) - g(x) + g(x) - f_N(x) + f_N(x) \\ \implies \abs{f_n(x)} \leq \abs{f_n(x) - g(x)} + \abs{g(x) - f_N(x)} + \abs{f_N(x)} \\ \leq \varepsilon + \varepsilon + M_N \]

by the above two statements. But \(N<\infty\), so we can choose \(M = \max\theset{M_1, M_2, \cdots M_{N-1}, 2\varepsilon + M_N}\) as a uniform bound. Then just take \(\varepsilon \to 0\). (Maybe not necessary?)

Goal: Let \(E\) be a set, and let \((f_n)_{n=1}^\infty\) be a sequence of functions \(f_n: E \to \mathbb{R}\) such that each \(f_n\) is bounded on \(E\). Suppose \(f_n \to g\) uniformly on \(E\). Prove that the sequence \((f_n)_{n=1}^\infty\) is uniformly bounded on \(E\), i.e., there exists a constant \(M > 0\) such that \(|f_n(x)| \le M\) for all \(n \in \mathbb{N}\) and all \(x \in E\).

<1>1. Definition of uniform convergence: There exists an integer \(N \ge 1\) such that for all \(n \ge N\) and all \(x \in E\), \(|f_n(x) - g(x)| < 1\).

<1>2. For all \(n \ge N\) and all \(x \in E\), \(|f_n(x)| \le |f_N(x)| + 2\). Proof: <2>1. By the triangle inequality: \[|f_n(x) - f_N(x)| \le |f_n(x) - g(x)| + |g(x) - f_N(x)|.\] <2>2. For any \(n \ge N\), applying <1>1 gives \(|f_n(x) - g(x)| < 1\) and \(|f_N(x) - g(x)| < 1\). Thus \(|f_n(x) - f_N(x)| < 1 + 1 = 2\). <2>3. Again by the triangle inequality: \[|f_n(x)| \le |f_N(x)| + |f_n(x) - f_N(x)| < |f_N(x)| + 2.\]

<1>3. For each \(k \in \{1, 2, \dots, N\}\), there exists a constant \(M_k < \infty\) such that \(|f_k(x)| \le M_k\) for all \(x \in E\).

<1>4. Define \(M = \max\{M_1, M_2, \dots, M_{N-1}, M_N + 2\}\). Then \(|f_n(x)| \le M\) for all \(n \in \mathbb{N}\) and all \(x \in E\). Proof: <2>1. If \(1 \le n < N\), then for all \(x \in E\), \(|f_n(x)| \le M_n \le M\) by <1>3 and definition of \(M\). <2>2. If \(n \ge N\), then by <1>2 and <1>3: \[|f_n(x)| < |f_N(x)| + 2 \le M_N + 2 \le M.\] <2>3. In all cases, \(|f_n(x)| \le M\) for all \(n \in \mathbb{N}\) and all \(x \in E\).

<1>5. Conclusion: \((f_n)_{n=1}^\infty\) is uniformly bounded on \(E\).

Uniform convergence to zero gives \[ \|f_n\|_{\infty}=\sup_{x\in[0,1]}|f_n(x)|\longrightarrow0. \] Therefore \[ \left|\int_0^1f_n(x)\,dx\right| \leq\int_0^1|f_n(x)|\,dx \leq\|f_n\|_{\infty}\longrightarrow0. \]

<1>1. For part (a), take \[ f(x)= \begin{cases} 0,&x\le0,\\ e^{-1/x},&x>0. \end{cases} \] This function is \(C^\infty\) on \(\mathbb R\), vanishes on \((-\infty,0]\), and is nonzero for every \(x>0\).