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Let $P, Q \colon \mathbb{R} \rightarrow \mathbb{R}$ are polynomials, and $Q(n) \neq 0$ for $n\in\mathbb{N}$ Suppose, that $\deg (P) < \deg (Q)$. Prove, that infinite sum



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Alternating series test? – Gerry Myerson Jan 6 '13 at 14:50
up vote 2 down vote accepted
  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 $P(x)= x^p+ax^{p-1}+\dots$ and $Q(x)=x^q+bx^{p-1}$. We have \begin{align} \frac{P(n)}{Q(n)}-n^{p-q}&=\frac{a n^{p+q-1}-dn^{p+q-1}+\dots}{n^{2q}}, \end{align} so we can find a constant $C$ such that for all $n$, $$\left|\frac{P(n)}{Q(n)}-n^{p-q}\right|\leqslant C\frac 1{n^{q-p+1}}\leqslant \frac C{n^2}.$$

  3. Now the only task is to show that for $p\geqslant 1$, the series $$\sum_{n=1}^{+\infty}\frac{(-1)^n}{n^p}$$ is convergent. Dirichlet criterion can be used, or an Abel transform. Let $s_n:=\sum_{k=0}^n(-1)^k$. We have \begin{align}\sum_{j=n+1}^{n+m}(-1)^j\frac{1}{j^p}&=\sum_{j=n+1}^{n+m}(s_j-s_{j-1})\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}{(m+n)^p}-s_n\frac{1}{n^p}+\sum_{j=n+1}^{n+m-1}s_j\left(\frac{1}{j^p}-\frac{1}{(j+1)^p}\right). \end{align} We conclude using the fact that $|s_j|\leqslant 1$ for all $j$.

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