Question #175369

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


1
Expert's answer
2021-03-29T05:26:18-0400

Let's show that

ϵ>0  ,n0=n0(ϵ)N:n,p>no    xn+pxn<ϵ\forall \epsilon >0 \;,\exists n_0 = n_0(\epsilon) \in N : \forall n,p > n_o \; \; |x_{n+p} -x_n| < \epsilon

xn+pxn=1(n+p)2+n+p+11n2+n+1<1(n+p)2+2(n+p)+11n2+2n+1=1(n+p+1)21(n+1)2\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,pNn, p \in N): n+p>n(n+p+1)2>(n+1)21(n+p+1)2<1(n+1)2n+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 ϵ>0\epsilon>0 xn+pxn<ϵ.|x_{n+p} - x_n|< \epsilon.

Thus, the given sequence is a Cauchy sequence.


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