 # Prove, that infinite sum sum_(n=1)^(infty)(-1)^n(P(n))/(Q(n)) Blaine Ortega 2022-08-13 Answered
Let $P,Q:\mathbb{R}\to \mathbb{R}$ are polynomials, and $Q\left(n\right)\ne 0$ for $n\in \mathbb{N}$ Suppose, that $\mathrm{deg}\left(P\right)<\mathrm{deg}\left(Q\right)$. Prove, that infinite sum
$\sum _{n=1}^{\mathrm{\infty }}\left(-1{\right)}^{n}\frac{P\left(n\right)}{Q\left(n\right)}$
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1. We can assume without loss of generality that P and Q are monic. Let p the degree of P, q those of Q (integers, if P=0, the problem is trivial), because the problem of convergence of the series is invariant up to a multiplication by a constant.
2. Write $P\left(x\right)={x}^{p}+a{x}^{p-1}+\dots$ and $Q\left(x\right)={x}^{q}+b{x}^{p-1}$. We have
$\begin{array}{rl}\frac{P\left(n\right)}{Q\left(n\right)}-{n}^{p-q}& =\frac{a{n}^{p+q-1}-d{n}^{p+q-1}+\dots }{{n}^{2q}},\end{array}$
so we can find a constant C such that for all n,
$|\frac{P\left(n\right)}{Q\left(n\right)}-{n}^{p-q}|⩽C\frac{1}{{n}^{q-p+1}}⩽\frac{C}{{n}^{2}}.$
3. Now the only task is to show that for $p⩾1$, the series
$\sum _{n=1}^{+\mathrm{\infty }}\frac{\left(-1{\right)}^{n}}{{n}^{p}}$
is convergent. Dirichlet criterion can be used, or an Abel transform. Let ${s}_{n}:=\sum _{k=0}^{n}\left(-1{\right)}^{k}$. We have
$\begin{array}{rl}\sum _{j=n+1}^{n+m}\left(-1{\right)}^{j}\frac{1}{{j}^{p}}& =\sum _{j=n+1}^{n+m}\left({s}_{j}-{s}_{j-1}\right)\frac{1}{{j}^{p}}\\ & =\sum _{j=n+1}^{n+m}{s}_{j}\frac{1}{{j}^{p}}-\sum _{j=n}^{n+m-1}{s}_{j}\frac{1}{{j}^{p}}\\ & ={s}_{m+n}\frac{1}{\left(m+n{\right)}^{p}}-{s}_{n}\frac{1}{{n}^{p}}+\sum _{j=n+1}^{n+m-1}{s}_{j}\left(\frac{1}{{j}^{p}}-\frac{1}{\left(j+1{\right)}^{p}}\right).\end{array}$
We conclude using the fact that $|{s}_{j}|⩽1$ for all j