# I am not able to prove this series

#### tarun

##### New member
(See attached pic)

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#### Jomo

##### Elite Member
Hi,
Why can't you do prove this? What is going wrong? What have you tried. It is hard to help you if we do not even know how you want to prove this. Please show us your attempts.
Personally I would try using mathematical induction

Is that it ?

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#### pka

##### Elite Member
Is that it ?
Well you have shown it is true for $$\displaystyle n=1$$.
Now for the hard part.
Suppose that $$\displaystyle \prod\limits_{k = 1}^n {\left( {\frac{{k\pi }}{{2n + 1}}} \right)} = \frac{1}{{{2^n}}}$$ is true.
On that basis you must show that $$\displaystyle \prod\limits_{k = 1}^{n+1} {\left( {\frac{{k\pi }}{{2[n+1] + 1}}} \right)} = \frac{1}{{{2^{n+1}}}}$$ is also true.

Thank you

#### Jomo

##### Elite Member
Is that it ?
If that is the case, then for example, 2 raised to any positive integer will be a single digit number because 21, 22 and 23 are all single digit numbers. Patterns do NOT always continue!