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1

$$e^{-1}+4e^{-9}+9e^{-25}+16e^{-49}\approx 0.36837308051278053458657911933771842$$ should be good enough. The truncation error is $$\sum_{i=4}^\infty(i+1)^2e^{-(2i+1)^2}=\sum_{i=4}^\infty(i+1)^2e^{-4i^2-4i-1}<\sum_{i=4}^\infty(i+1)^2e^{-64-4i-1}<2\cdot10^{-34}$$ as can be computed analytically. You can obtain this estimate from $$\sum_{i=n}^\infty ...


2

$$\sum_{s\geq 1}\frac{s}{(s-1)!}\lambda^{s-1} = \sum_{n\geq 0}\frac{n+1}{n!}\lambda^{n}=\frac{d}{d\lambda}\left(\lambda e^{\lambda}\right)=\color{red}{(\lambda+1)\,e^{\lambda}}.$$


2

Hint $$\sum_{s=1}^\infty \frac{\lambda^{s-1}}{(s-1)!}s=\frac{\mathrm d }{\mathrm d \lambda}\sum_{s=1}^\infty \frac{\lambda ^s}{(s-1)!}$$


1

$$\frac{\lambda^{s-1}}{(s-1)!}s=\frac{\lambda^{s-1}}{(s-1)!}(s-1)+\frac{\lambda^{s-1}}{(s-1)!}$$ Can you go on?


1

Write that sum as $S(a,t)$. Here are some thoughts: If $\gcd(a,p)=1$, then $S(a,t)=S(1,t)$ because raising a primitive $p^t$-root of unity to $a$ still gives you a primitive $p^t$-root of unity. If $\gcd(a,p)>1$, then $S(a,t)=m S(a',t')$ for some $m\in \mathbb N$ and $a'<a$ with $\gcd(a',p)=1$ and $t'<t$. So, the only sum that you need to care ...


2

I'm not sure that this exactly what you are looking for, but here comes anyway. The additive group of $F[x]$, $F=\Bbb{F}_q$, $q=p^n$ is just an infinite dimensional vector space over the prime field $\Bbb{F}_p$. Therefore the values of all characters are $p$th complex roots of unity. A popular (standard?) description the characters of the additive group $...



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