$$\sum_{n=1}^\infty \frac{3^x(n+2)}{n!}, x\in E = (0 , +\infty)$$ It is likely easy, but for me it is new thing and I don't know how to start to solve this. Could you give me any tips?

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    $\begingroup$ Are you sure? $$\sum_{n=1}^\infty \frac{3^x(n+2)}{n!}=3^x\sum_{n=1}^\infty \frac{(n+2)}{n!}$$ $\endgroup$ – Aforest Nov 5 '16 at 3:59
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    $\begingroup$ Agree with @Aforest. One you've taken $3^x$ outside the sum, think about the sum as a Taylor series (or the sum of two Taylor series) and put it into closed form. Uniform convergence is not really an issue here, it's just a sequence convergence independent of $x$. $\endgroup$ – Scott Burns Nov 5 '16 at 4:15
  • $\begingroup$ @ScottBurns You have described convergence, not uniform convergence. Can you show that for all $\epsilon>0$, there is a number $N_0$ independent of $x$, such than whenever $N>N_0$, $$3^x\sum_{n=N}^\infty\frac{n+2}{n!}<\epsilon?$$ $\endgroup$ – Mark Viola Nov 5 '16 at 4:42
  • $\begingroup$ @Dr.MV You're right ! The range of the function $3^x$ must be specified as bounded. The result holds for any fixed bounded domain $D\subset \mathbb{R}$ or $D\subset \mathbb{C}$. $\endgroup$ – Duchamp Gérard H. E. Nov 5 '16 at 4:45
  • $\begingroup$ Note the sequence $3^x s_n$ is uniformly convergent in $x\in(0,\infty)$ if and only if $s_n$ is constant for $n$ sufficiently large, as $3^x$ is unbounded on $(0,\infty). $\endgroup$ – user251257 Nov 5 '16 at 4:45

There exists a number $\epsilon=1$ such that for all $N$, there exists a number $x_N=\log_3((N+2)!)$ and there exists a number $N+1>N$ such that

$$\begin{align} 3^{x_N}\left|\sum_{n=1}^\infty \frac{n+2}{n!}-\sum_{n=1}^{N+1}\frac{n+2}{n!}\right|&=3^{x_N}\sum_{n=N+2}^\infty\frac{n+2}{n!}\\\\ &\ge \frac{3^{x_N}}{(N+2)!}\\\\ &=1 \\\\ &=\epsilon \end{align}$$

This is precisely the negation of uniform convergence.


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