A problem on Mathematical Induction

Question

Prove that for each odd natural number $n\geq3$

$(1+\frac{1}{2})(1-\frac{1}{3})(1+\frac{1}{4})............(1+\frac{(-1)^n}{n})=1$

By mathematical Induction

we write the given series is $\Pi_{k=2}^n (1+\frac{(-1)^n)}{n})=1$

for n=3 , LHS=$\frac{3}{2}\frac{2}{3}=1$, RHS=$1$

Hence LHS=RHS..true for n=3

suppose this is true for for some odd n

ie

$(1+\frac{1}{2})(1-\frac{1}{3})(1+\frac{1}{4})............(1+\frac{(-1)^n}{n})=1$

now we have to prove for n+2

consider $(1+\frac{1}{2})(1-\frac{1}{3})(1+\frac{1}{4})............(1+\frac{(-1)^n}{n})(1+\frac{(-1)^{n+2}}{n+2})$

how we processed next

Answer 1

It's certainly true for $n=3$. $3/2 \cdot 2/3 = 1$. Try working out the next example, $n=5$. You should notice an interesting cancellation. Does it keep happening for $n=7,9,11,\ldots $?

Answer 2

We would like to prove $$ \prod_{k=2}^n\left(1+\frac{(-1)^k}{k}\right)=1 $$ for odd $n\geq 3$. As already pointed out, this is indeed true for $n=3$. For the induction step note that for odd $n$ we have \begin{align} \left(1+\frac{(-1)^{n+1}}{n+1}\right)\left(1+\frac{(-1)^{n+2}}{n+2}\right)&=\frac{n+2}{n+1}\cdot\frac{n+1}{n+2} =1. \end{align}