Past exam of the mathematics course of the University of Cambridge / 2010 / ia / Paper 1 / 4E (Analysis I)

Pairing the first and last terms of the arithmetic progression gives ${\sum }_{k=1}^{n}k=n(n+1)/2$. Therefore the limit of a sequence is

The sequence is at least $1$. Fix $\epsilon >0$. By the binomial theorem, for $n\ge 2$,once $(n-1){\epsilon }^2>2$. Taking positive $n$th roots gives $1\le n^{1/n}<1+\epsilon$ for all sufficiently large $n$. This proves directly from the definition of a limit of a sequence that

Since $0<a\le b$, we have $b^n\le a^n+b^n\le 2b^n$. Taking positive $n$th roots yieldsFor every $\epsilon >0$, the binomial theorem gives $(1+\epsilon {)}^n\ge 1+n\epsilon >2$ eventually, so $2^{1/n}\to 1$. The squeeze theorem therefore gives