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3

To get the value, split the sum, use the Taylor series for $\ln(1+x)$ (also known as Mercator series) and the alternating harmonic series: $$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}(1+2^n)}{n2^n} =\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n2^n}+\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}= \ln\left(1+\frac{1}{2}\right)+\ln 2= \ln 3$$

2

Hint. You may write $$\frac{nx^n}{4n^2-1}=\frac{x^n}{4 (2 n-1)}+\frac{x^n}{4 (2 n+1)}$$ and one may recall $$\sum_{n=0}^\infty\frac{u^{2n+1}}{2 n+1}=\frac12 \log\left(\frac{1+u}{1-u}\right), \qquad |u|<1.$$

2

Hint: $\text{arctanh }x=\displaystyle\sum_{n=1}^\infty\frac{x^{2n-1}}{2n-1}$

2

$$\frac{x^{n-1}}{2n-1} = \frac{y^{2n-2}}{2n-1} \quad \text{where }y=\sqrt x$$ and $$\frac d {dy}\, \frac{y^{2n-1}}{2n-1} = y^{2(n-1)} = x^{n-1}.$$ $$\sum_{n=1}^\infty \frac{x^{n-1}}{2n-1} = \sum_{n=1}^\infty \frac{y^{2n-1}}{2n-1}.$$ The derivative of this with respect to $y$ is $$\sum_{n=1}^\infty y^{2n-1} = \frac y {1-y^2} = \frac A {1-y} + \frac B ... 2 Taking the 40th derivative at 0 will yield you the coefficient for the x^{40} term multiplied by 40! The general series for natural log is:$$ ln(x+1) = \frac {x^1}1 - \frac {x^2}2 + \frac {x^3}3 - \frac {x^4}4... $$replacing x^2 makes:$$ ln(x^2+1) = \frac {x^2}1 - \frac {x^4}2 + \frac {x^6}3 - \frac {x^8}4... $$From this it is apparent that ... 2 As Daniel fischer noted this argument only works for the special case of non negative terms b_i . Hint: if the set \{n: b_n \geq 1\}  is infinite then you should know what to do. However if that is finite,then try to make use of the following fact: for every n  such that b_n<1 we have$$\frac {b_n}{1+b_n} > \frac {b_n}{2} $$1 Hint: For the case F(0)=0 (if the function F is not 0) you can write F(x)=x^mG(x) for a G such that G(0)\not =0(and of radius of convergence R). Hence you can find an \varepsilon, 0<\varepsilon<R, such that if |x|\leq \varepsilon, we have$$c=\frac{|G(0)|}{2}\leq |G(x)|\leq \frac{3|G(0)|}{2} =d$$Now in ... 1 A Taylor series about x=a is given by$$f(x) = \sum_{n=0}^\infty \frac{ f^{(n)}(a)}{n!} (x-a)^n$$where f^{(n)}(a) is the nh derivative of f evaluated at x=a. The 0th derivative is just the function itself. To actually find the series you can try to compute the first several terms given from the series above and see if you discover a pattern. ... 1 Integrating you find that$$ f(z)=(1-z)\log(1-z)+z. $$It cannot be extended as an analytic (or even continuous) function on a neighborhood of z=1, which is a branching point. However f is continuous on \{|z|\le1\} (when defined at z=1 as f(1)=1.) 1 f(x)=\frac{x^7}{a^8-x^8}=\frac{x^7}{a^8}\frac{1}{1-(\frac{x}{a})^8}=\frac{x^7}{a^8}\sum_{i=0}^\infty(\frac{x}{a})^{8i} with convergence interval as |\frac{x}{a}|<1 1 As (1-x^k)-(1-x^{k+1})=x^k(x-1),$$\frac{x^k}{\left(1-x^k\right) \left(1-x^{k+1}\right)}=\dfrac{(1-x^k)-(1-x^{k+1})}{(x-1)(1-x^k)(1-x^{k+1})}=\dfrac1{x-1}\left(\dfrac1{1-x^{k+1}}-\dfrac1{1-x^k}\right)$$Can you recognize the telescoping nature? 1$$a_k=\frac{x^k}{(1-x^k)(1-x^{k+1})}=\frac{1}{1-x^k}-\frac{1}{1-x^{k+1}}+\frac{x^{k+1}}{(1-x^k)(1-x^{k+1})}\\ =\frac{1}{1-x^k}-\frac{1}{1-x^{k+1}}+x\cdot a_k\sum_{k=1}^{n}a_k=\frac{1}{1-x}-\frac{1}{1-x^{n+1}}+x\sum_{k=1}^{n}a_k\sum_{k=1}^{n}a_k=\frac{\frac{1}{1-x}-\frac{1}{1-x^{n+1}}}{1-x}$$1 Given R=\lim\limits_{n\to\infty} \frac{a_n}{a_{n+1}}=\lim\limits_{n\to\infty} \frac{a_{n-1}}{a_{n}} Also, let$$\sum \limits_{n=0}^{\infty} \frac{a_n x^{n+1}}{n+1}=\sum \limits_{n=1}^{\infty} \frac{a_{n-1} x^{n}}{n}=\sum \limits_{n=1}^{\infty} b_{n} x^{n}$$where b_n=\frac{a_{n-1}}{n} R_1=\lim\limits_{n\to\infty} ... 1 Outline: Our function can be rewritten as$$f(x)=\frac{x^3}{2^3}\cdot \frac{1}{(1-x/2)^3}.$$To find the expansion of \frac{1}{(1-t)^3}, note that for suitable t we have$$\frac{1}{1-t}=1+t+t^2+t^3+\cdots. Differentiate twice with respect to $t$.

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