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## Hot answers tagged power-series

8

For every $n$, $$S_n=\sum_{k=1}^n\frac1k2^k\geqslant\sum_{k=1}^n\frac1n2^k=\frac1n(2^{n+1}-1).$$ On the other hand, for every $u$ in $(0,1)$, $$S_n=\sum_{k\lt un}\frac1k2^k+\sum_{un\leqslant k\leqslant n}\frac1k2^k\leqslant\sum_{k\lt un}2^k+\sum_{un\leqslant k\leqslant n}\frac1{un}2^k\leqslant2^{un+1}+\frac1{un}2^{n+1}.$$ This holds for every $u\lt1$ hence ...

6

Truncating series is a good way to hunt for roots, you just have to do it carefully, as Micah points out. I'll show that there are at least $10$ roots in the unit disc. I would guess there are infinitely many. Set $$f(z) = z+z^2+z^4+z^8+\cdots \quad \mbox{and} \quad f_0=z+z^2+z^4+\cdots+z^{256}$$ According to Mathematica, $|f_0(z)| > 0.066$ on ...

4

The Euler-Maclaurin summation formula is useful for approximating sums and often reveals the asymptotic behavior with only a few terms. This problem is an interesting application because the precise asymptotic behavior requires summing an infinite number of terms with Bernoulli numbers as coefficients - the terms that are typically neglected. Using the ...

3

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.

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 $$\lim_{n \to \infty} 2^n+1 = \lim_{n \to \infty} ... 3 It is the definition of a special function called trilogarithm \operatorname{Li}_3(x). This is a special case of the polylogarithm$$\operatorname{Li}_n(x)=\sum_{k=1}^{\infty}\frac{x^k}{k^n},$$which has the properties x\operatorname{Li}'_n(x)=\operatorname{Li}_{n-1}(x) and \operatorname{Li}_n(1)=\zeta(n). These identities naturally generalize those ... 3 Hint: You might consider first integrating parts:$$\begin{align} \int_{0}^{1/2}x^3\arctan{(x)}\,\mathrm{d}x &=\frac14x^4\arctan{(x)}\bigg{|}_{0}^{1/2}-\frac14\int_{0}^{1/2}\frac{x^4}{1+x^2}\,\mathrm{d}x\\ &=\frac{1}{64}\arctan{\left(\frac12\right)}-\frac14\int_{0}^{1/2}\frac{x^4}{1+x^2}\,\mathrm{d}x\\ ...

3

This is an extended comment to the answer of Semiclassical. The series $$f(z)=\sum_{k=0}^{\infty}z^{2^k}$$ is a canonical example of a function having natural boundary (the so-called lacunary series). Here the natural boundary is given by the unit circle $|z|=1$, inside which the series is absolutely convergent. A more general statement is the ...

3

This answer is closer to a sketch than a proof. If you notice gaps/mistakes/nonsense in this presentation, please let me know... We want to examine the behavior of the function $f(z)$ inside the square root. It evidently vanishes linearly at the origin. Indeed, for all $|z|<1$ this series is convergent by comparison with the geometric series. If ...

3

Don't let the $(-1)^k$ or $(-x)^k = (-1)^kx^k$ trouble you. They have the effect of canceling each other out for odd $k$, and besides, for the ratio test, we apply it taking the absolute value of the general term $|a_k|$. $$|a_k| = \frac{(x)^k }{k}$$ $$\frac{a_{k+1}}{a_k} = \frac{\frac{(x)^{k+1}}{k+1}}{\frac{(x)^k }{k}} = \frac{xk}{k+1}$$

3

No, it is possible that the radii of convergence become arbitrarily small. We can write down explicit examples, e.g. $$f(z) = \frac{1}{\sin \left(\pi(z^2-i)\right)}$$ or $$g(z) = \frac{1}{\sin \left(\pi(z^2-i)\right)} + \frac{1}{\sin \left(\pi (z^2+i)\right)}$$ if you prefer functions that are real-valued on $\mathbb{R}$. More generally, every domain ...

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

So your power series, for part (a)., we must conduct a test (Ratio test or Root test) to find the interval of values of x such that the series converges. So, using the root test: $$\lim_{n\to\infty} \left|\frac{x^n}{e^{\sqrt{\ln n}}}\right|^{1/n}$$ $$= \lim_{n\to\infty} \left|\frac{x}{e^{\frac{\sqrt{\ln n}}{n}}}\right|$$ The numerator of the fraction ...

2

We can use a standard real example. If we want a non-real example, multiply everything by $i$. Consider the series $$1+\frac{1}{2}z+\frac{1}{3}z^2+\frac{1}{4}{z^3}+\frac{1}{5}z^4+\cdots.$$ This converges for $z=-1$, but does not converge absolutely there, since the harmonic series $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots$ diverges. We can replace ...

2

Your series for $e^x$ and $\cos(x^2)$ are correct. If you want to multiply two series, think about $(1+a_1x+a_2x^2+a_3x^3+\dots )(1+b_1x+b_2x^2+b_3x^3+\dots )=1+(a_1+b_1)x+(a_2+b_2+a_1b_1)x^2+\dots$ Can you see where each term in $x, x^2$ comes from? Can you do the $x^3$ term?

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

You already know that $$\log(1-x)=-\sum_{k=1}^{\infty} \frac{x^k}{k}=\sum_{k=1}^{\infty} a_kx^k$$ Then, $$a_k=-\frac{1}{k}$$ The ratio test, then, is: $$\biggl|{\frac{a_{k+1}}{a_k}}\biggr|=\frac{\frac{1}{k+1}}{\frac{1}{k}}=\frac{k}{k+1}$$ The convergence radius $R$ is given by: $$\lim_{k\rightarrow \infty}\biggl|\frac{a_{k+1}}{a_k}\biggr|=\frac{1}{R}$$ ...

2

By the Cauchy-Hadamard formula; the radius of convergence of a power series isgiven by $$\frac1{R}=\limsup |a_n|^{1/n}$$ Now , we may plug in and see that $|n^n|^{1/n!} \to 1$ as $n\to \infty$ .So we conclude that $R=1$.

2

Yes you are allowed to do that. The power series $$(1 + A)^{\frac12} = \sum_{n=0}^\infty \frac{(-1)^n(2n)!}{(1-2n)(n!)^2(4^n)}A^n = 1 + \textstyle \frac{1}{2}A - \frac{1}{8}A^2 + \frac{1}{16} A^3 - \frac{5}{128} A^4 + \dots,$$ converges at least in the region $\|A\|<1$. The convergence is locally uniform and "absolute", but the term "absolute" should ...

2

$F_2$ is the sum of the odd powers of $x$ divided by factorials, it should remind you of the Taylor development of $e^x$. Indeed, you get the series with only odd terms by combining $e^x$ and $e^{-x}$: $$F_2(x)=\frac{e^x-e^{-x}}2=\sinh x.$$ This is known as the hyperbolic sine. Similarly, $F_1$ is the sum of even powers, the hyperbolic cosine, but all ...

2

First of all, as yoyo says, these can be solved through differential equations. For example, if $y=2+F_1(x)$, then we have $$y''=y,y(0)=2,y'(0)=0$$ However, given both $F_1(x)$ and $F_2(x)$ to work with, we can use $e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$ to see that $$\frac{F_1(x)}2+F_2(x)+1=e^x$$ $$\frac{F_1(x)}2-F_2(x)+1=e^{-x}$$ From here, it's simple ...

2

If $\alpha=p/q$ and $n\ge q$ then $\alpha n!$ is an integer and so $$(re^{2\pi i\alpha})^{n!}=r^{n!}\ .$$ So the tail of the sum is $$r^{q!}+r^{(q+1)!}+r^{(q+2)!}+\cdots\ ;$$ this diverges if $|r|\ge1$. It converges if $|r|<1$, but I don't know of any simple expression for the sum.

2

$${(-1)}^{n-1}\frac{1}{n!{x}^{n}}=-\frac{\left(-\frac1x\right)^n}{n!}$$ $$\implies\sum_{n=0}^{\infty}{(-1)}^{n-1}\frac{1}{n!{x}^{n}}=-\sum_{n=0}^{\infty}\frac{\left(-\frac1x\right)^n}{n!}=-e^{-\frac1x}$$ $$\implies(-1)^{0-1}\frac1{0! x^0}+\sum_{n=1}^{\infty}{(-1)}^{n-1}\frac{1}{n!{x}^{n}} =-e^{-\frac1x}$$ ...

2

Note that $(-1)^n(-1)^n=(-1)^{2n}=1$ and you end up with the well-known (and divergent) harmonic series.

2

When $x=-1$: $$\sum_{n=0}^{\infty} \frac{(-1)^n x^n}{n+1}=\sum_{n=0}^{\infty} \frac{(-1)^n (-1)^n}{n+1}=\sum_{n=0}^{\infty} \frac{(-1)^{2n} }{n+1} \\ =\sum_{n=0}^{\infty} \frac{1 }{n+1} \text{, it is the harmonic series,and we know that this diverges.}$$ When $x=1$: $$\sum_{n=0}^{\infty} \frac{(-1)^n x^n}{n+1}=\sum_{n=0}^{\infty} \frac{(-1)^n ... 2 It suffices to show that the coefficients a_n cannot converge to 0. Now suppose that |a_k| \leq \varepsilon for k \geq m. Then for z\in\mathbb{D}$$|f(z)|\leq \frac{\varepsilon}{1-|z|}+\sum_{k=0}^{m-1}|a_k|.$$This implies that if a_k converges to 0 then \lim_{|z|\to 1}(1-|z|)|f(z)|=0. In particular f cannot have a pole on \mathbb{T} ... 2 This is an expansion and follow up of my comment. As mentioned in comment. \hspace0.5in Without further restriction, there are no relation. An example is the function f(x) = x^3 which is invertible over the real axis and yet its inverse function doesn't have a power series expansion at x = 0. In general, if your function is invertible only over ... 2 With d'Alembert (I think, it's also known as "Ratio Test"):$$\lim_{n\to\infty }\frac{\left|\frac{(-1)^{n+1}}{(n+1)!}\right|}{\left|\frac{(-1)^n}{n!}\right|}=\lim_{n\to\infty }\frac{n!}{(n+1)!}=\lim_{n\to\infty }\frac{1}{n+1}=0$$Then the radius of convergence is \mathcal R=\infty . 2 The matrix$$ A(x):=\sum_{n=0}^\infty A_n x^n $$is invertible at x=1. The set of invertible matrices is an open set in the space of all matrices. Thus, an idea to solve the problem is to show that x\mapsto A(x) is continuous. In order to prove that you need additional assumptions. One would be to assume that the power series$$ \sum_{n=0}^\infty ...

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