The inductive proof of $\sum_{k=1}^n \frac k{k+1} \leq n - \frac1{n+1}$ is unclear

Remember that in induction proofs, we start by assuming that the claim we're trying to prove is true for $n$, and then conclude that it must also be true for $n + 1$. In this case, we start with the induction assumption $$\sum_{k = 1}^{n} \frac{k}{k + 1} \leq n - \frac{1}{n + 1},$$ and want to end with the conclusion that $$\sum_{k = 1}^{n + 1} \frac{k}{k + 1} \leq n + 1 - \frac{1}{n + 2} .$$ Here's the argument written out in a bit more detail with commentary on each step. \begin{align*} \sum_{k = 1}^{n + 1} \frac{k}{k + 1} & = \frac{n + 1}{n + 2} + \sum_{k = 1}^{n} \frac{k}{k + 1} & \textrm{ (just writing out the sum)} \\ & \leq \frac{n + 1}{n + 2} + n - \frac{1}{n + 1} & \textrm{ (applying the induction hypothesis)} \\ & = 1 - \frac{1}{n + 2} + n - \frac{1}{n + 1} & \textrm{ (rewriting $\frac{n+ 1}{n + 2}$ as $\frac{n + 2 - 1}{n + 2} = 1 - \frac{1}{n + 2}$)} \\ & = n + 1 - \frac{1}{n + 1} - \frac{1}{n + 2} & \textrm{ (regrouping)} \\ & = n + 1 - \frac{(n + 2) + (n + 1)}{(n + 2)(n + 1)} & \textrm{ (combining fractions)} \\ & = n + 1 - \frac{2(n + 2) - 1}{(n + 1)(n + 2)} & \textrm{ (regrouping the numerator)} \\ & = n + 1 - \frac{2(n + 2)}{(n + 1)(n + 2)} + \frac{1}{(n + 1)(n + 2)} & \textrm{ (breaking the fraction back apart)} \\ & = n + 1 - \frac{2}{n + 1} + \frac{1}{(n + 1)(n + 2)} & \textrm{ (simplifying the fraction)} \\ & \leq n + 1 - \frac{2}{n + 2} + \frac{1}{(n + 1)(n + 2)} & \textrm{ (we slightly modified the second-to-last summand)} \\ & \leq n + 1 - \frac{2}{n + 2} + \frac{1}{n + 2} & \textrm{ (modifying the last summand)} \\ & = n + 1 - \frac{1}{n + 2} & \textrm{ (combining the fractions)} . \end{align*} So in summary, we began with the assumption that the claim held for $n$, and through some arithmetic trickery concluded that it therefore held for $n + 1$.

If $a\leq b $, then $a+c\leq b+c $ for any $c\in \Bbb R $. By assumption, we have $$\sum_{k=1}^{n}\frac k{k+1}\leq n-\frac 1{n+1}.$$ Now add $\dfrac{n+1}{n+2}$ on both sides, i.e., $$\sum_{k=1}^{n}\frac k{k+1}+\frac{n+1}{n+2}\leq n-\frac 1{n+1}+\frac{n+1}{n+2}.$$ Note that $$\sum_{k=1}^{n}\frac k{k+1}+\frac{n+1}{n+2}=\sum_{k=1}^{n+1}\frac k{k+1}.$$ Hence we have $$\sum_{k=1}^{n+1}\frac k{k+1}\leq n-\frac 1{n+1}+\frac{n+1}{n+2}.$$