1. ## Riemann Integrals, Limits

Prove that $\displaystyle a_n$ has a limit:

$\displaystyle a_n=\sum_{i=1}^{2^n} sin\frac{i}{2^{2n}}cos\frac{i}{2^{2n}}cos\frac{i}{ 2^{2n-1}}...cos\frac{i}{2^{n+1}}$

The subject is integrals, Riemann's definition and all that... I have no idea how to look for a solution

Thanks!

Prove that $\displaystyle a_n$ has a limit:

$\displaystyle a_n=\sum_{i=1}^{2^n} sin\frac{i}{2^{2n}}cos\frac{i}{2^{2n}}cos\frac{i}{ 2^{2n-1}}...cos\frac{i}{2^{n+1}}$

The subject is integrals, Riemann's definition and all that... I have no idea how to look for a solution

Thanks!
Step 1::

Take d = i/2^(n+1)

Use 2 sin a . cos a = sin 2a

For example,

2 sin i/2^k . cos i/2^k = sin i/2^(k-1)

Step 2 ::

Then use Comparision Test with series 1/2^(n+1)

3. Originally Posted by ques
Step 1::

Take d = i/2^(n+1)

Use 2 sin a . cos a = sin 2a

For example,

2 sin i/2^k . cos i/2^k = sin i/2^(k-1)

Step 2 ::

Then use Comparision Test with series 1/2^(n+1)
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Well.. it looks like an optimal solution, but I am looking for a solution that involves use of integrals, and the Riemann's definition of integrals. Something like $\displaystyle \sum_{i=1}^n f(x_i)\Delta x_i$