Tag Info

Hot answers tagged

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 ...



Only top voted, non community-wiki answers of a minimum length are eligible