2007 H2 Mathematics Paper 2 Question 2

Sigma Notation

Answers

(i)

Out of syllabus

(ii)

1−1(N+1)2{1 - \frac{1}{(N+1)^2}}

(iii)

As N→∞,{ N \to \infty, } 1(N+1)2→0{\frac{1}{(N+1)^2} \to 0} so 1−1(N+1)2→1{1-\frac{1}{(N+1)^2} \to 1}
Hence the series is convergent
Sum to infinity =1{= 1}

(iv)

1−1N2{1-\frac{1}{N^2}}

Full solutions

(i)

Out of syllabus

(ii)

∑n=1N2n+1n2(n+1)2=∑n=1N(un−un+1)=u1−u2+u2−u3+u3−u4+⋯+uN−2−uN−1+uN−1−uN+uN−uN+1=u1−uN+1=1−1(N+1)2  ■\begin{align*} & \sum_{n=1}^N \frac{2n+1}{n^2(n+1)^2} \\ & = \sum_{n=1}^N (u_n - u_{n+1}) \\ & = \def\arraystretch{1.5} \begin{array}{lclc} & u_1 &-& \cancel{u_2} \\ + & \cancel{u_2} &-& \cancel{u_3} \\ + & \cancel{u_3} &-& \cancel{u_4} \\ + & & \cdots & \\ + & \cancel{u_{N-2}} &-& \cancel{u_{N-1}} \\ + & \cancel{u_{N-1}} &-& \cancel{u_{N}} \\ + & \cancel{u_{N}} &-& u_{N+1} \\ \end{array} \\ &= u_1 - u_{N+1} \\ &= 1 - \frac{1}{(N+1)^2} \; \blacksquare \end{align*}

(iii)

As N→∞,{ N \to \infty, } 1(N+1)2→0{\displaystyle \frac{1}{(N+1)^2} \to 0} so
∑n=1N2n+1n2(n+1)2=1−1(N+1)2→1\begin{align*} & \sum_{n=1}^N \frac{2n+1}{n^2(n+1)^2} \\ &= 1 - \frac{1}{(N+1)^2} \\ &\to 1 \end{align*}
Hence the series is convergent ■{\blacksquare}
Sum to infinity=∑n=1∞2n+1n2(n+1)2=1  ■\begin{align*} & \textrm{Sum to infinity} \\ &= \sum_{n=1}^\infty \frac{2n+1}{n^2(n+1)^2} \\ &= 1 \; \blacksquare \end{align*}

(iv)

Replacing n{n} with n+1,{n+1,}
∑n=2N2n−1n2(n−1)2=∑n=1N−12(n+1)−1(n+1)2(n+1−1)2=∑n=1N−12n+1(n+1)2n2=1−1(N−1+1)2=1−1N2  ■\begin{align*} & \sum_{n=2}^N \frac{2n-1}{n^2(n-1)^2} \\ & = \sum_{n=1}^{N-1} \frac{2(n+1)-1}{(n+1)^2(n+1-1)^2} \\ & = \sum_{n=1}^{N-1} \frac{2n+1}{(n+1)^2n^2} \\ & = 1 - \frac{1}{(N-1+1)^2} \\ & = 1 - \frac{1}{N^2} \; \blacksquare \end{align*}