Find the formula for the following summary: $$\sum_{k=1}^{n}(\dfrac{1}{2k}-\dfrac{1}{2k+1}+\dfrac{1}{k+n})$$
Click here for the solution.Solution: The formula is $$ S(n)=\sum_{k=1}^{n}(\dfrac{1}{2k}-\dfrac{1}{2k+1}+\dfrac{1}{k+n})=\dfrac{2n}{2n+1}$$
Proof by induction: For $n=1$, $$S(1)=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{2}{3}=\dfrac{2\cdot 1}{2\cdot 1+1}$$ and the claim is correct.
Assume for $n=m$, the claim is correct, therefore $$S(m)=\dfrac{2m}{2m+1}$$ Then $$ S(m+1)=S(m)-\dfrac{1}{m+1}+(\dfrac{1}{2(m+1)}-\dfrac{1}{2(m+1)+1}+\dfrac{1}{2(m+1)})+\dfrac{1}{2m+1}$$ $$ = S(m)-\dfrac{1}{2m+3}+\dfrac{1}{2m+1}=\dfrac{2m}{2m+1}+\dfrac{2}{(2m+1)(2m+3)}$$ $$ = \dfrac{2m(2m+3)+2}{(2m+1)(2m+3)}=\dfrac{2(m+1)(2m+1)}{(2m+1)(2m+3)}=\dfrac{2(m+1)}{2(m+1)+1}$$
Therefore, for $n=m+1$, the claim is correct, and the formula is proved by induction.
Proof by expansion: By expanding the summary, it can be shown that all fractions are cancelled out, except the following $$ \dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+…+\dfrac{1}{2^m}+\dfrac{1}{2^m}-\dfrac{1}{2n+1}=1-\dfrac{1}{2n+1}=\dfrac{2n}{2n+1}$$ where $2^m$ is the largest power of 2 less or equal to $n$.