Instructions
Click the button below to start.
Parts (i)-(iii) will stay the same. Part (iv) will be changed each time you click the button again.
Practice until you can perform the "substitution" technique reliably.
Question
A sequence $u_n$ is defined by $ \displaystyle u_n = \frac{1}{n^3}$ for $n \in \mathbb{Z}, n \geq 1$.
- Show that $\displaystyle u_n - u_{n+1} = \frac{3n^2+3n+1}{n^3(n+1)^3}$.
- Hence find $\displaystyle \sum_{n=1}^N \frac{3n^2+3n+1}{n^3(n+1)^3}$.
- Give a reason why the series in part (ii) is convergenet and state the sum to infinity.
- Use your answer to part (ii) to find $$1$$
Answer
- $\displaystyle 1 - \frac{1}{(N+1)^3}$.
- As $N \to \infty$, $\displaystyle \frac{1}{(N+1)^3} \to 0$. $1 - \displaystyle \frac{1}{(N+1)^3} \to 1$ so the series is convergent and the sum to infinity is $1$.
- $$2$$