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3

Let $\omega=\exp\left(\frac{2\pi i}{3}\right)$. Then $\frac{\omega^n+\omega^{2n}+1}{3}$ equals $1$ iff $3\mid n$, hence: $$ \sum_{n\geq 1}\frac{x^{3n}}{(3n)!} = \color{red}{-1+\frac{e^{x}+e^{\omega x}+e^{\omega^2 x}}{3}}. \tag{1}$$


3

Define $$y(x)=\sum_{k\geq0}\frac{x^{3k}}{\left(3k\right)!}. $$ We observe that $$\sum_{k\geq0}\frac{x^{3k}}{\left(3k\right)!}+\sum_{k\geq0}\frac{x^{3k+1}}{\left(3k+1\right)!}+\sum_{k\geq0}\frac{x^{3k+2}}{\left(3k+2\right)!}=e^{x} $$ so we have the second order ODE $$y''\left(x\right)+y'\left(x\right)+y\left(x\right)=e^{x}.\tag{1} $$ Let start to solve the ...


2

To expound my comment, Consider: $$ S(x,t) = \sum_{1 \le n \le x} e^{\frac{\pi}{2}xn^2}\left(e^{ia(t)}\left(\frac{\pi}{2}xn^2 \right)^{3/4+it}+e^{-ia(t)}\left(\frac{\pi}{2}xn^2 \right)^{3/4-it} \right) $$ Where $ a(t)=arg(\Gamma(1/4+it)) $. Then: $$ S(x,t) = \sum_{1 \le n \le x} e^{\frac{\pi}{2}xn^2}\left(\frac{\pi}{2}xn^2 ...


2

If $0<R<\infty$, then the series $\sum_{n=0}^{\infty}R^{-n}x^n$ has radius of convergence $R$, because for fixed $x$ it is just a geometric series.


2

If you have power series $\sum_{n\geq0}a_nq^n$ and $\sum_{n\geq0}b_nq^n$, and you know that they are equal, then you know that $a_n=b_n$ for all $n\geq0$. That is «equating the like powers of $q$» in the two series.


1

Since $f$ has only finitely many singularities $z_1,……,z_m$ which are simple poles, then $f$ can be extended to some $B(0,R)(R>1)$ to become a meromorphic functions. Without lost of generality we set the non-zero residue $Res(f,z_j)=c_j$, then $$g(z):=f(z)-\sum_{j=1}^m \frac{c_j}{z-z_j} \in H(B(0,R))$$ and set $$f(z)=\sum_{n=0}^{\infty}a_n z^n,z \in ...


1

Sure there is: If $$ \sum_{n=0}^{n=\infty}c_n(z-a)^n$$ is a power series with radius of convergence R, then for any $\lambda > 0$ the series $$\sum_{n=0}^{n=\infty}c_n\left(\frac{z-a}{\lambda}\right)^n$$ converges iff $\left\lvert\frac{z-a}{\lambda}\right\rvert < R$, i.e., iff $|z-a| < \lambda R$. Hence, the power series ...


1

The following assumes that you already know $e^x = \sum_{n=0}^\infty x^n/n! $, ie the Taylor series of $e^x $ at $0$ (and that it converges to $e^x $). Given this, simply note $$ \sum e (x-1)^n /n! = e \cdot \sum (x-1)^n/n! = e \cdot e^{x-1} = e^x. $$


1

We need to show that $\displaystyle\lim_{n\to\infty} R_n(x)=0$ for all x, where $\displaystyle R_n(x)=\frac{f^{n+1}(c)(x-1)^{n+1}}{(n+1)!}=\frac{e^c(x-1)^{n+1}}{(n+1)!}$ and $c$ is a number between 1 and $x$. 1) If $x< 1$, then $\displaystyle \lim_{n\to\infty}R_n(x)=0$ since $\displaystyle |R_n(x)|< \frac{e|x-1|^{n+1}}{(n+1)!}$ (since $e^c< e$) ...



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