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

We have that $(1 + {1 \over 2})(1 - {1 \over 3}) = 1 - {1 \over 3} + {1 \over 2} - {1 \over 6} = 1$. So the formula is true for $n=3$. The very first term starts from $(1 + {1 \over 2})$. Likely you started with $(1 - {1 \over 3})$ which is not correct. — Decaf-Math, Oct 17 '16 at 04:14
The thing that you did wrong here is that you left off the $n+1$ factor of the product in your induction step (note that the index is from $k=2$ to $n+2$). — rnrstopstraffic, Oct 17 '16 at 05:09

score 2 · Answer 1 · answered Oct 17 '16 at 04:21

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}

answered Oct 17 '16 at 04:21

sranthrop

8,497

@sranthrop...thanks,so u proved for n-1 and assumed for n=k+1 and then we proved k=n+2 – Murali Oct 17 '16 at 04:47
Uhm no. Once you have the statement settled for some odd $n$, you can use "my" induction step to prove $n+2$ (which is the next odd number). – sranthrop Oct 17 '16 at 04:49
@sranthrop..your series running from n=2...can you give me brief solution please – Murali Oct 17 '16 at 04:50
@sranthrop..yes i understand why your series running from k=2 to n... – Murali Oct 17 '16 at 04:54
Just write down the induction step to show us that you have understood what to do (just edit your question for that). :) – sranthrop Oct 17 '16 at 04:55
...i edited my attempt can you help me for next – Murali Oct 17 '16 at 05:03
@sranthrop...i edited ca you now help me solve this problem – Murali Oct 17 '16 at 05:05
Ok nice. We see now where your mistake was made: You are missing the term $(1+\frac{(-1)^{n+1}}{n+1})$ in your induction step! If you add this term, the first $n$ terms factor to 1 (induction!), and the last two terms can be tackled using "my" induction step. – sranthrop Oct 17 '16 at 05:06
@sranthrop....sorry,,but thanks for your time – Murali Oct 17 '16 at 05:10
You're welcome. Did you get in now? – sranthrop Oct 17 '16 at 05:19
@sranthrop...i got it – Murali Oct 17 '16 at 05:20

score 1 · Answer 2 · answered Oct 17 '16 at 04:14

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 $?

answered Oct 17 '16 at 04:14

Reese Johnston

17,988

Not a negative comment, but for those who are having difficulty seeing what this answer means in the first line, try writing in the extra step: $$\left(1 + {1\over2}\right)\left(1 - {1 \over 3}\right) = \left({3 \over 2}\right)\left({2 \over 3}\right) = {6 \over 6} = 1.$$ – Decaf-Math Oct 17 '16 at 04:22

A problem on Mathematical Induction

2 Answers2