Question: Prove that harmonic numbers satisfy the equality using induction $$ H_{1}+ H_{2} + · · · + H_{n} = (n + 1)H_{n} − n. $$ I have done the basis step: $(1 + 1)H_{1} − 1 = 1$. Correct.
Done the inductive hypothesis: $(k + 1)H_{k} − k$ and started solving: $$ (k + 2)H_{k+1} − (k+1) = (k + 1)H_{k} − k + \frac{1}{k+1}. $$ What do I do once I get to this part?
You’ve made a mistake on the righthand side of what you’re trying to prove: if
$$H_1+H_2+\ldots+H_k=(k+1)H_k-k\;,$$
then
$$H_1+H_2+\ldots+H_k+H_{k+1}=(k+1)H_k-k+H_{k+1}\;,$$
not
$$(k+1)H_k-k+\frac1{k+1}\;.$$
Thus, you should be trying to prove that
$$(k+2)H_{k+1}-(k+1)=(k+1)H_k-k+H_{k+1}\;.$$
Approach it the same way you’d approach proving a trigonometric identity: algebraically manipulate one side into the other, or both into an identical third expression. All you really need is that $H_{k+1}=H_k+\frac1{k+1}$.