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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 $f(1/2)$.

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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 ...

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

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(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 $a$.)

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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 ...

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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}.$$In ... 1 The intuitive geometric approach is to look at all the nth roots. I mean, draw them as points on a piece of paper (or in your head). See how nicely they're distributed, all evenly along the unit circle. That means their sum is equal to 0. Now raise them to some power. If that power is a multiple of n, then they will all end up at 1, and their sum ... 1 If |z|<1, then \left| \frac{z^n}{n^2} \right|\leq \frac{1}{n^2} =: M_n and \sum_{n=1}^{\infty} M_n < \infty. Therefore, the Weiestrass M-Test implies that the series \sum_{n=1}^{\infty} \frac{z^n}{n^2} converges uniformly on \{ z \in \mathbb{C} \colon |z| < 1 \} 1 Let S be the sum. If we multiply S by 1/2 and subtract from S, we have$$ 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 ...

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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 ...

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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 ...

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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 ...

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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 $x<0$.

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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.

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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

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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 ...

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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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