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Solutions to Real Analysis: A Long-Form Mathematics Textbook Chapter 1


Exercise 1.10

Solution: We remark the following identity\[\frac{1}{k}-\frac{1}{k+1}=\frac{(k+1)-k}{k(k+1)}=\frac{1}{k(k+1)}.\]Therefore, we have\begin{align*}&\quad\sum_{k=1}^{n}\frac{1}{k(k+1)}=\sum_{k=1}^{n}\left(\frac{1}{k}-\frac{1}{k+1}\right)\\&=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots+\left(\frac{1}{n-1}-\frac{1}{n}\right)+\left(\frac{1}{n}-\frac{1}{n+1}\right)\\&=1-\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}-\frac{1}{3}\right)-\cdots-\left(\frac{1}{n}-\frac{1}{n}\right)-\frac{1}{n+1}\\&=1-\frac{1}{n+1}=\frac{n}{n+1}\end{align*}


Another solution: We prove it by induction on $n$.

If $n=1$, the identity is clearly true.

Suppose the identity is true for $n$, we prove it for $n+1$. By induction hypothesis, we have\begin{equation}\label{anal1.10.1}\sum_{k=1}^{n}\frac{1}{k(k+1)}=\frac{n}{n+1}.\end{equation}Then we have\begin{align*}\sum_{k=1}^{n+1}\frac{1}{k(k+1)}&=\sum_{k=1}^{n}\frac{1}{k(k+1)}+\frac{1}{(n+1)(n+2)}\\\text{use }\eqref{anal1.10.1}\quad&=\frac{n}{n+1}+\frac{1}{(n+1)(n+2)}=\frac{n(n+2)+1}{(n+1)(n+2)}\\&=\frac{n^2+2n+1}{(n+1)(n+2)}=\frac{(n+1)^2}{(n+1)(n+2)}\\&=\frac{n+1}{n+2}.\end{align*}Hence the identity is true for $n+1$.

By Mathematical Induction, we are done!

Linearity

This website is supposed to help you study Linear Algebras. Please only read these solutions after thinking about the problems carefully. Do not just copy these solutions.

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