Answer to Question #175369 in Real Analysis for Anand

Question

Show that 1/(n²+n+1) n belongs to N is a Cauchy sequence

Answer 1

Let's show that

$\forall \epsilon >0 \;,\exists n_0 = n_0(\epsilon) \in N : \forall n,p > n_o \; \; |x_{n+p} -x_n| < \epsilon$

$\displaystyle |x_{n+p} -x_n|= \frac{1}{(n+p)^2+n+p+1} - \frac{1}{n^2+n+1}< \frac{1}{(n+p)^2+2(n+p)+1} - \frac{1}{n^2+2n+1} = \frac{1}{(n+p+1)^2} - \frac{1}{(n+1)^2}$

for any n and p ($n, p \in N$): $n+p>n \Rightarrow (n+p+1)^2 > (n+1)^2 \Rightarrow \frac{1}{(n+p+1)^2} < \frac{1}{(n+1)^2}$

Then, for any $\epsilon>0$ $|x_{n+p} - x_n|< \epsilon.$

Thus, the given sequence is a Cauchy sequence.