Riemann Integrals, Limits

**adam63** · May 5th 2010, 06:09 AM

Prove that $a_n$ has a limit:

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

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

Thanks!

**ques** · May 5th 2010, 02:59 PM

Originally Posted by adam63

Prove that $a_n$ has a limit:

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

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)

**adam63** · May 6th 2010, 02:18 AM

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)

---

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 $\sum_{i=1}^n f(x_i)\Delta x_i$

**adam63** · May 7th 2010, 09:41 PM

Can anyone please help me with that? I need it urgently.

**Drexel28** · May 7th 2010, 10:07 PM

Originally Posted by adam63

Can anyone please help me with that? I need it urgently.

Let's see some work. What does the summand simplify to?

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