# Power series $\sum_{n=0}^\infty \frac{2n+3}{(2n)!}t^{2n}$

\begin{align*} \sum_{n=0}^\infty \frac{2n+3}{(2n)!}t^{2n}&= \sum_{n=0}^\infty \frac{2n}{(2n)!}t^{2n}+ 3\sum_{n=0}^\infty \frac{t^{2n}}{(2n)!}=\left\{\begin{array}{c} 2n=k\\ n=0\Rightarrow k=0\\ n=\infty \Rightarrow k=\infty\end{array} \right\}\\ &=\sum_{k=0}^\infty \frac{k}{k!}t^{k}+ 3\sum_{k=0}^\infty \frac{t^{k}}{k!}=t\cdot e^t+3 e^t, \end{align*}

where $r=0$ is convergence radius, so the series converges only for $t=0$.

Is this correct?

• Looks like the series is convergent for all $t$, not just $t=0$. Also, $k$ is always even, so the sums are not $e^t$ and $te^t$. – Quang Hoang Apr 12 '16 at 7:01
• I thought of that also, but for $\sin t$ and $\cos t$ I require $(-1)^n$, right? – zermelovac Apr 12 '16 at 7:02

$$e^x=\sum_{r=0}^\infty\dfrac{x^r}{r!}$$
$$e^x+e^{-x}=?$$
$$e^x-e^{-x}=?$$
$$\dfrac{2n}{2n!}t^{2n}=t\dfrac{t^{2n-1}}{(2n-1)!}$$
• $e^x+e^{-x}=\sum_{r=0}^\infty \left( \frac{x^r}{r!}+(-1)^r \frac{x^r}{r!}\right)=\sum_{r=0}^\infty (1+(-1)^r) \frac{x^r}{r!}=\sum_{r=0}^\infty2 \frac{x^{2r}}{(2r)!} \Rightarrow \sum_{r=0}^\infty \frac{x^{2r}}{(2r)!}=cosh x$ Similar, $\sum_{r=0}^\infty \frac{x^{2r+1}}{(2r+1)!}=sinh x$... Calculating I get, $\sum_{r=0}^\infty x \frac{x^{2r-1}}{(2r-1)!}=x\cdot sinhx$ – zermelovac Apr 12 '16 at 7:23