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8

Using the binomial theorem and the Cauchy convolution we have: $$\begin{eqnarray*}[z^h]\left(\frac{1+z}{1-z}\right)^{1/3}&=&(-1)^h\sum_{l\leq h}\binom{-1/3}{l}\binom{1/3}{h-l}(-1)^l\\ &=& (-1)^h\binom{-1/3}{h}\cdot\phantom{}_2 F_1\left(-\frac{1}{3},-h;\frac{2}{3}-h,-1\right).\end{eqnarray*}\tag{1}$$ Since: $$\lim_{h\to+ \infty}\phantom{}_2 ...


7

Note that $$\frac{1}{i!}\left(\frac{e^2-1}{2}\right)^i \prod_{j=0}^i (x-2j) = x \prod_{j=1}^i \frac{\frac{x}{2}-j}{j} (e^2-1)^i = x\binom{\frac{x}{2}-1}{i}(e^2-1)^i,$$ so $$\sum_{i=0}^\infty \frac{1}{i!}\left(\frac{e^2-1}{2}\right)^i \prod_{j=0}^i (x-2j) = x\sum_{i=0}^\infty \binom{\frac{x}{2}-1}{i}(e^2-1)^i = x\left(1+(e^2-1)\right)^{\frac{x}{2}-1} = x ...


5

Hint: For which values of $y$ does $$ \sum_{n=1}^{\infty} y^n $$ converge? Now let $$ y = \frac{1}{1+|z|^2}. $$


5

Hint Start with $$\frac{1}{1-x}=\sum_{i=0}^{\infty}x^i$$ and integrate with respect to $x$. You then have $$\log(1-x)=-\sum_{i=1}^{\infty}\frac{x^i}{i}$$ So, $$x^n \log(1-x)=-\sum_{i=1}^{\infty}\frac{x^{n+i}}{i}$$ Added later to this answer If you want to start with the derivative and then integrate, there is no problem. If $$f(x)=x\log(1-x)$$ ...


4

Let $w = \frac{z+1}{z-1}$. Then you have a power series in $w$, centered at $0$. Find its radius of convergence, call that $R$. Then find which $z$ correspond to $\lvert w\rvert < R$. The map $z \mapsto \frac{z+1}{z-1}$ can be explicitly inverted.


4

Write $$ \left(\dfrac{1+x}{1-x}\right)^{1/3} = \frac{(1+x)^{1/3} - 2^{1/3}}{(1-x)^{1/3}} + \frac{2^{1/3}}{(1-x)^{1/3}} =: f(x)+g(x). $$ The function $f$ is continuous on the disk $|z| \leq 1$, so by Darboux's method (Thm. VI.14 in Analytic Combinatorics) we have $$ [x^n] f(x) = o\left(\frac{1}{n}\right). $$ The coefficients of the second series are given ...


4

If $x\leqslant1$, then $a_n\geqslant1$ for every $n$ hence the series $\sum\limits_na_n$ diverges. If $x\gt1$, then the expansion $x^{1/n}=\exp((\log x)/n)=1+(\log x)/n+o(1/n)$ yields $$\log(a_{n+1}/a_n)=\log(2-x^{1/n})\sim-(\log x)/n,$$ hence $\log a_n\sim-(\log x)\cdot\log n$. Thus: If $x\gt\mathrm e$, one can pick $1\lt y\lt\log x$ then ...


4

You essentially solved it; just make the substitution $$ f(x) = \frac{1}{2+3x} = \frac{1}{2}\left(\frac{1}{1-\left(\frac{-3x}{2}\right)} \right) = \frac{1}{2}\sum_{n=0}^\infty \left(\frac{-3x}{2}\right)^n. $$ Note that $1/(1-x)$ converges when $|x| < 1$, though, so in this problem the radius of convergence is $$ \left|\frac{-3x}{2}\right| < 1 \implies ...


3

I would really like to know the closed form given in the book. The following answer was given for the original post. The only thing I have been able to do to arrive to something is to replace $n!$ by Stirling approximation, that is to say $$n! \simeq \sqrt{2 \pi } e^{-n} n^{n+\frac{1}{2}}$$ and using this, the result of the summation is given by ...


3

A rigorous way to define the sine function is to consider it as the solution to the IVP: $$ \begin{cases} y^{\prime \prime} + y = 0\\ y(0) = 0 \\ y^{\prime}(0) = 1 \end{cases} $$


3

$$\lim_{n \to \infty}\sqrt[n]{\frac{2^n+1}{n}} = \lim_{n \to \infty}\sqrt[n]{\frac{2^n}{n}} =\lim_{n \to \infty}\frac{2}{ \sqrt[n]{n}}=2$$ Why can we omit the $+1$ term? Good question. ;) Write $$2^n+1 = 2^n(1+\frac{1}{2^n})$$ As $n \to \infty$, the limit of both sides becomes equal. Therefore $$\lim_{n \to \infty} 2^n+1 = \lim_{n \to \infty} ...


3

By the root test we have $$\left|2^n\frac{(4z-8)^n}n\right|^{1/n}\xrightarrow{n\to\infty}2|4z-8|<1\iff|z-2|<\frac18\iff z\in B\left(2,\frac18\right)$$ so the radius is $\frac18$.


3

You can't apply the limit test because you don't know the limits $|\frac{a_n}{a_{n+1}}|$ or $|\frac{b_n}{b_{n+1}}|$. If these limits exist, you can use them to find a RoC, but knowing the radius doesn't imply that the limits exist.* To show the answer is $3$, recall that a power series converges absolutely (and uniformly) inside it's radius of convergence. ...


3

Note that $S'(x)=\displaystyle\sum\limits_{k=0}^\infty \frac{(-1)^k x^{2k}}{(2k)!}$. The first term here is $1$, so when you differentiate, it dies. Then $$S''(x)=\sum_{k\geqslant 1}\frac{(-1)^k x^{2k-1}}{(2k-1)!}$$ i.e. we start the sum at $k=1$. Can you continue? All you need to do now is shift an index. Note that we repeatedly used $(x^n)'=nx^{n-1}$ and ...


2

For one direction, if $f$ has a removable singularity or a pole of order $k$ in $0$, what can you say about the function $z\mapsto z^m f(z)$ for large enough $m$? For the other direction, note that if $f$ has an essential singularity in $0$, then so has $z \mapsto z^m f(z)$ for all $m \in \mathbb{Z}$. Appeal to the Casorati-WeierstraƟ theorem if necessary.


2

Hint: Note that $\displaystyle\sum_{j=0}^{\infty}(j-2)a_{j}x^{j}=(x\frac d{dx}-2)\sum_{i=0}^{\infty}a_{i}x^{i}$


2

As told in answers and comments, a series exists but it is quite tedious to derive. Writing $$\frac{\displaystyle\sum_{i=0}^{\infty}a_{i}x^{i}}{\displaystyle\sum_{j=0}^{\infty}(j-2)a_{j}x^{j}}=\sum_{i=0}^{\infty}b_{i}x^{i}$$ the first coefficients are given by $$b_0=-\frac{1}{2}$$ $$b_1=-\frac{a_1}{4 a_0}$$ $$b_2=\frac{a_1^2-4 a_0 a_2}{8 a_0^2}$$ ...


2

You have $\displaystyle G(x)=(1+x+x^2+x^3)^5=\big(\frac{1-x^4}{1-x}\big)^5=(1-x^4)^5(1-x)^{-5}$ $\displaystyle=\big(1-5x^4+\binom{5}{2}x^8-\binom{5}{3}x^{12}+\cdots\big)\big(\sum_{k=0}^{\infty}\binom{k+4}{4}x^4\big)$, so now you just have to find the coefficient of $x^{12}$ in this expression.


2

Some hints: When $\sum_{k=1}^\infty c_k$ converges then $\sum_{k=1}^\infty c_k\>\rho^k$ converges absolutely for every positive $\rho<1$. Distinguish the cases $|z|<1$ (easier case) and $|z|>1$. Solution: When $\sum_{k=1}^\infty c_k$ converges then there is an $M>0$ with $|c_k|\leq M$ for all $k\geq1$. Assume $|z|=:\rho<1$. As ...


2

Hint: Use $$ 2 - \sin x \leq 3 $$ and then compare with a geometric series.


2

Here is a variant of this question asked on MathOverflow: http://mathoverflow.net/q/49395/. In particular, your question is answered by the cited result that $D$ can be any $G_\delta$ subset of the circle (such a set can be uncountable with dense complement).


2

The thought process should probably begin with: let's look at some function with a simple pole on the unit circle, and with a power series that I can find. This should remind of the example $$\frac{1}{1-z} = \sum_{n=0}^\infty z^n$$ Here the coefficients are all equal to $1$, so they are bounded. More generally, we can place a simple pole at any point $a$ ...


2

Hint. Check that $$\sum_{k=0}^{n} \binom{n}{k} (-1)^{k} \frac{n^{2}+n+1}{nk+n+1} = \prod_{k=1}^{n} \left( 1+\frac{1}{nk} \right)^{-1}. $$ The starting idea is to write $$\sum_{k=0}^{n} \binom{n}{k} (-1)^{k} \frac{n^{2}+n+1}{nk+n+1} = (n^{2}+n+1) \sum_{k=0}^{n} \binom{n}{k} (-1)^{k} \int_{0}^{1} x^{nk+n} \, dx. $$


2

Your ratio is incorrect. You should have: $$\frac{c_n}{c_{n+1}}=\frac{2^{3n}(n+1)}{2^{3(n+1)}n}=\frac{2^{3n}(n+1)}{2^{3n+3}n}=\frac{n+1}{2^3n}$$


2

Hint This is not really a question about complex numbers. Think geometric series. For the question as it stands now, you do not have to use the convergence radius expression. But to answer your question about $a_n$, if you will use the Ratio Test then you should use $a_n=\frac{1}{(1+|z|^2)^n}$. Note that our series is not a power series in the usual sense. ...


2

In general, to get the power series (or Taylor series at $x=0$) for $f(g(x))$ we need the Taylor series for $g(x)$ around $0$ and the Talyor series of $f(x)$ around $g(0)$. In the case of $\sqrt{\cos x}$, that means $f(x)=\sqrt{x}$ needs a Taylor series for $\sqrt{x}$ around $x=1$, which is: $$\sqrt{x} = \sum_{k=0}^\infty \binom{1/2}{k} (x-1)^k$$ Even if ...


2

Some other important formulas regarding $\sin(x)$ that you didn't mention are the infinite product $$\sin(x) = x \prod_{k=1}^{\infty}\Big( 1 - \frac{x^2}{\pi^2 k^2} \Big)$$ and the partial fractions decomposition $$\frac{1}{\sin(x)^2} = \sum_{k=-\infty}^{\infty} \frac{1}{(x-\pi k)^2}, \; \; x \notin \pi \cdot \mathbb{Z},$$ although I guess the latter only ...


2

HINT: Use binom expansion: $$(1+a)^{y}=1+a.y+\frac{a^2}{2!}.y(y-1)+\frac{a^3}{3!}.y(y-1)(y-2)+....$$ $y=\frac{x}{2}$ $$(1+a)^{\frac{x}{2}}=1+a.\frac{x}{2}+\frac{a^2}{2!}.\frac{x}{2}(\frac{x}{2}-1)+\frac{a^3}{3!}.\frac{x}{2}(\frac{x}{2}-1)(\frac{x}{2}-2)+....$$ ...


2

It suffices to show that the coefficients $a_n$ cannot converge to $0$. Now suppose that $|a_k| \leq \varepsilon$ for $k \geq m$. Then for $z\in\mathbb{D}$ $$|f(z)|\leq \frac{\varepsilon}{1-|z|}+\sum_{k=0}^{m-1}|a_k|.$$ This implies that if $a_k$ converges to $0$ then $\lim_{|z|\to 1}(1-|z|)|f(z)|=0$. In particular $f$ cannot have a pole on $\mathbb{T}$ ...


2

The given series is convergent by the Leibniz's theorem but it isn't absolutely convergent since the harmonic series is divergent hence this series is conditionally convergent.



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