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7

Put $$f(y)=\sum_{x=1}^{\infty} xy^x=y\sum_{x=1}^{\infty} xy^{x-1}=y\left(\sum_{x=1}^{\infty} y^x\right)'=y\left(\frac{1}{1-y}\right)'$$ Compute the right-hand side above and then put $f(1/2)$.


6

Just to see how ugly it is, you actually can do it using Cauchy products: \begin{align*}\cos^2 x&=\left(\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}\right)^{\!2}=\sum_{n=0}^\infty\sum_{k=0}^n (-1)^{n-k}\frac{x^{2(n-k)}}{(2(n-k))!}(-1)^k\frac{x^{2k}}{(2k)!}=\sum_{n=0}^\infty\sum_{k=0}^n(-1)^n\binom{2n}{2k}\frac{x^{2n}}{(2n)!}\\ \sin^2 ...


4

$$x^2e^{-x}=x^2\sum_{n=0}^\infty\frac{(-1)^nx^n}{n!}=\sum_{n=0}^\infty\frac{(-1)^nx^{n+2}}{n!}\implies$$ $$2xe^{-x}-x^2e^{-x}=\sum_{n=0}^\infty (-1)^n\frac{(n+2)x^{n+1}}{n!}$$ ...and the above is true for all $\;x\in\Bbb R\;$ .


3

(Assuming that $\tau$ in your formula stands for $2 \pi$) this is an immediate consequence of the "well-known" formula for the (finite) geometric sum $$ \sum_{k=0}^{n-1} x^k = \cases {n \quad \quad \text{ if } x = 1\\ \frac{x^n-1}{x-1}\quad \text{ otherwise}} $$ applied to $x = e^{\tau ia/n}$. (Note that $x = 1$ exactly if $n$ divides $a$.)


3

HINT: $$(n+1)(n-1)=n^2-1=n(n-1)+n-1$$ $$\implies\frac{(n+1)(n-1)}{n!}x^n=\frac{n(n-1)+n-1}{n!}x^n =x^2\cdot\frac{x^{n-2}}{(n-2)!}+x\cdot\frac{x^{n-1}}{(n-1)!}-\frac{x^n}{n!}$$ Now $\sum_{r=0}^\infty\dfrac{x^r}{r!}=e^x$ More generally for $a_0+a_1n+a_2n^2+\cdots$ in the numerator, we can set this to $a_0+b_1n+b_2n(n-1)+\cdots$ Now set $n=1,2$ etc. to ...


3

The power series of the exponential function is defined on $\Bbb R$ so we can differentiate it term by term on $\Bbb R$ and we get $$\exp'(x)=\exp(x)$$ Moreover, we see easily that $\exp(x)>0$ for $x\ge0$ and using the Cauchy product we get $$\exp(x)\exp(y)=\exp(x+y),\quad \forall (x,y)\in\Bbb R^2$$ hence $$\exp(-x)\exp(x)=\exp(0)=1\implies ...


2

Yes we can integrate term by term a power series on its domain of convergence so in your case $$\sum_{n=0}^\infty (-1)^n x^n$$ is a power series and its domain of convergence is $(-1,1)$ so for all $x\in(-1,1)$ we have $$\ln(1+x)=\int_0^x\frac{dt}{t+1}=\int_0^x\sum_{n=0}^\infty (-1)^n t^ndt=\sum_{n=0}^\infty(-1)^n\frac{x^{n+1}}{n+1}$$


2

Since the series $$\sum_{n\ge0} a_n$$ is divergent then the Radius $R_a$ of the given power series is less or equal $1$. Moreover since $a_n\le A_n$ then $R_a\ge R_A$ where $R_A$ is the radius of convergence of $$\sum_{n\ge0}A_nx^n$$ Finally we have $$\frac{A_{n}}{A_{n+1}}=1-\frac{a_{n+1}}{A_{n+1}}\xrightarrow{n\to\infty}1$$ so by the ratio criteria we ...


2

Assuming the $a_n$'s are non-negative (or that we have absolute convergence at $x_0$), then notice that $|a_n e^{-\lambda_n x }| \leq a_n e^{-\lambda x_0}$ for any $x \in [x_0,\infty)$. Since $\sum_{n=0}^{\infty} a_n e^{-\lambda_n x_0} < \infty$, the Weierstrass M-Test implies that $\sum_{n=0}^{\infty} a_n e^{-\lambda_n x}$ converges uniformly for $x \in ...


2

Justify that you can differentiate the series term by term and find, for all $x\in \mathbb R$, $$\sin'(x)=\sum \limits_{n=0}^\infty\left(\dfrac{(-1)^nx^{2n}}{(2n)!}\right)=\cos(x).$$ The first equality is straightforward, I can't imagine what you're missing. Let me know if you need it and I'll try to clarify. Similarly $\cos'=-\sin$. To prove that ...


2

This doesn't use the Cauchy product formula, but you can resolve this identity using the power series themselves. Using the power series we find that $$\frac{d}{dx}\sin(x) = \cos(x)$$ and $$\frac{d}{dx} \cos(x) = -\sin(x).$$ Note also that: $$2\sin(x)\cos(x)-2\cos(x)\sin(x)=0$$ Integrating both sides gives us $$(\sin(x))^2 + (\cos(x))^2 = C$$ for some ...


2

By definition, $\displaystyle\sum_{n=-\infty}^\infty r^{n^2}=\theta_3(0,r)$. See Jacobi elliptic $\theta$ function.


2

One way to do this is to use the Newton generalized binomial theorem $$ (1-2x)^{-5} = \sum_{k=0}^{\infty} {-5 \choose k }(-1)^k (2x)^k $$ which gives you $$ [x^4](1-2x)^{-5} = (-1)^4 2^4{-5 \choose 4 }=1120. $$


1

Something that you might find helpful are Blaschke products. If a sequence of numbers $|z_n| < 1$ satisfy the condition $\sum_{n=0}(1-|z_n|) < \infty$ then there is a function analytic in the disc for which $f(z_n)=0$. In particular, the Blaschke product is such a function $$B(z)=\prod_{n=0}^\infty \frac{|z_n|}{z_n} \frac{z_n - z}{1-\bar{z_n}z}.$$ In ...


1

The intuitive geometric approach is to look at all the $n$th roots. I mean, draw them as points on a piece of paper (or in your head). See how nicely they're distributed, all evenly along the unit circle. That means their sum is equal to $0$. Now raise them to some power. If that power is a multiple of $n$, then they will all end up at $1$, and their sum ...


1

If $|z|<1$, then $\left| \frac{z^n}{n^2} \right|\leq \frac{1}{n^2} =: M_n$ and $\sum_{n=1}^{\infty} M_n < \infty$. Therefore, the Weiestrass M-Test implies that the series $\sum_{n=1}^{\infty} \frac{z^n}{n^2}$ converges uniformly on $\{ z \in \mathbb{C} \colon |z| < 1 \}$


1

Let $S$ be the sum. If we multiply $S$ by $1/2$ and subtract from $S$, we have $$ S/2 = (1\cdot (1/2) + 2\cdot (1/2)^2 +\ldots)-(0\cdot (1/2) + 1\cdot(1/2)^2 + 2\cdot (1/2)^3+\ldots)$$ Regrouping terms based on the power of $(1/2)$, we have $$ S/2 = (1-0)(1/2)+(2-1)(1/2)^2+(3-2)(1/2)^3+\ldots$$ and so $$S=1+1/2+1/4+1/8+\ldots = 2.$$


1

Hint: Solving your recurrence relationship gives $$ a_n=c_12^{-n}+c_2n2^{-n}, $$ where $c_1$ and $c_2$ can be determined by $a_0$ and $a_1$. You will end up calculating the sums $$ \sum_{n=0}^{+\infty} \Bigl(\frac{x}{2}\Bigr)^n $$ and $$ \sum_{n=0}^{+\infty} n\Bigl(\frac{x}{2}\Bigr)^n $$ Do you know how to do that? Ask to fill in details where necessary... ...


1

Since the characteristic polynomial of the recurrence $$ a_{n+2}=a_{n+1}-\frac{1}{4}a_n \tag{1}$$ is $p(x) = x^2-x+\frac{1}{4} = \left(x-\frac{1}{2}\right)^2$, we have that $$\left(2-x\right)^2\, f(x) = ax+b\tag{2}$$ where $a,b$ depend on the initial conditions. By decomposing $\frac{ax+b}{(2-x)^2}$ into partial fractions, we get: $$ a_n = 2^{-(n+2)} (b+(2 ...


1

Write \begin{align} \sum_{n=0}^{\infty}\frac{(n-1)(n+1)}{n!}x^n &=-1+\sum_{n=2}^{\infty}\frac{(n-1)n}{n!}x^n+\sum_{n=1}^{\infty}\frac{(n-1)}{n!}x^n\\ &=-1+x^2\sum_{n=2}^{\infty}\frac{(n-1)n}{n!}x^{n-2}+x\sum_{n=1}^{\infty}\frac{n}{n!}x^{n-1}-\sum_{n=1}^{\infty}\frac{x^{n}}{n!} \end{align} The first series in the sum is the second derivative of the ...


1

No, this is in general not true. Let $D = (-1,1)^2$, let $\Omega=\mathbb{R} \times (-1,1)$, both viewed as subsets of the complex plane, and let $\phi:D \to \Omega$ be the conformal map with $\phi(0)=0$, $\phi'(0)>0$, so that $\phi((-1,1)) = \mathbb{R}$. Using this conformal map, we can translate your problem into a similar problem on $D$, as follows: If ...


1

If $x>0$, you have a series with positive terms, so its sum is greater than the sum of the first two terms, which is $1+x$ and $1+x>1$. If $x<0$ you have an alternating series. It's a theorem that for alternating series, the error bound when you take the sum up to rank $m$: $\sum_{k=0}^m a_k$, is at most $\lvert a_{m+1}\rvert$ and the error has the ...


1

For $x>0$, $$\exp(x)=1+x+\frac{x}{2}+\cdots>1+0+0+\cdots=1$$ Since $\exp(0)=1$ and $\exp$ is strictly monontone increasing, $\exp(x)<1$ for $x<0$.


1

The note says that if $|z| \leq 1$, then we have absolute convergence, since the sum of the probabilities is bounded above by 1. Therefore, the radius of convergence is at least 1, hence $r_X \geq 1$. We do not have enough information to conclude how much bigger (if at all) the radius of convergence is.


1

Since your question isn't clear, I'll assume the sum is from $k = 0$ to $\infty$ $$\sum_{k\ge 0} kx^{-k} = -x \sum_{k\ge 0} (-k)x^{-k-1} = -x\;\frac{d}{dx}\! \left(\sum_{k\ge 0} x^{-k}\right) = -x \;\frac{d}{dx}\left(\frac{1}{1-x^{-1}} \right)$$ You do the rest


1

We use information that you may already know about geometrically distributed random variables. The mean waiting time until the first T is $2$. After that, the mean additional waiting time until the first H is $2$. And $2+2=4$. Remark: There are many other approaches, including series. An interesting alternate approach uses conditioning on the result of the ...


1

Let $N$ denote the number of tosses until you see “TH” for the first time. Find $\mathsf E(N)$ In order to encounter the termination event we must toss a series of none or more heads, a series of one or more tails, and then one head. The pattern, $\mathsf{[T]_XH[H]_YT}$ Here $X$ and $Y$, the number of tosses of a given face before the opposite face have ...



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