Infinite Series

**Hweengee** · October 28th 2008, 07:44 AM

Need help with the following:

Evaluate $\sum_{n=0}^{\infty }\sin \left( {\frac {1}{2^n}} \right) \cos \left( {\frac {3}{2^n}} \right)$

I recognise that the terms inside the sin and cos functions are the terms of a geometric progression with first term 1 and 3 respectively and common ratio of 0.5. But when it comes to evaluating the series with the trigonometric functions applied i'm quite lost. Is there some kind of general principle to adhere to to tackle problems of this nature?

Determine if $\sum _{n=0}^{\infty } \left( {\frac {3n}{4\,n+1}} \right) ^{2\,n}$ is convergent or divergent.

I'm abit confused as to which test to use for convergence/divergence for this particular problem. I'm thinking either limit or limit comparison, but I still can't seem to work it out. Would appreciate any help you guys might be able to offer.

Thanks in advance.

**ThePerfectHacker** · October 28th 2008, 09:29 AM

Originally Posted by Hweengee

Need help with the following:

Evaluate $\sum_{n=0}^{\infty }\sin \left( {\frac {1}{2^n}} \right) \cos \left( {\frac {3}{2^n}} \right)$

Remember that $2\sin A\cos B = \sin (A+B) + \sin (A-B)$

Therefore,
$2\sum_{n=0}^{\infty} \sin \left( {\frac {1}{2^n}} \right) \cos \left( {\frac {3}{2^n}} \right) = \sum_{n=0}^{\infty} \sin \left( \frac{1}{2^{n-2}} \right) - \sin \left( \frac{1}{2^{n-1}}\right) = \sin 4$

**Hweengee** · October 28th 2008, 10:36 AM

Thanks! And would it be correct to say that for the 2nd series,
$ \sum _{n=0}^{\infty } \left( {\frac {3n}{4\,n+1}} \right) ^{2\,n} $
is convergent

since
$ \lim _{n\rightarrow \infty}\left({\frac {3n}{4\,n+1}}\right)^{2\,n}=0 $

**ThePerfectHacker** · October 28th 2008, 11:24 AM

Originally Posted by Hweengee

$ \lim _{n\rightarrow \infty}\left({\frac {3n}{4\,n+1}}\right)^{2\,n}=0 $

Use the root test. Notice we get $\left( \frac{3n}{4n+1} \right)^2 \to \left( \frac{3}{4} \right)^2 < 1$ .
Thus, it is convergent.

**Mathstud28** · October 28th 2008, 12:38 PM

Originally Posted by Hweengee

Thanks! And would it be correct to say that for the 2nd series,
$ \sum _{n=0}^{\infty } \left( {\frac {3n}{4\,n+1}} \right) ^{2\,n} $
is convergent

since
$ \lim _{n\rightarrow \infty}\left({\frac {3n}{4\,n+1}}\right)^{2\,n}=0 $

Or you could alternatively say that since

$\lim_{n\to\infty}\sqrt[n]{\left|\left(\frac{3n}{4n+1}\right)^{2n}\right|}=\ left|\frac{9}{16}\right|<1$

Therefore it is convergent by the root test

EDIT: sorry...I noticed that hacker already did it this way...you could also see that

$0\leq\left(\frac{3n}{4n+1}\right)^{2n}\leq\left(\f rac{3n}{4n}\right)^{2n}=\left(\frac{9}{16}\right)^ n$

A convergent geometric series

**ThePerfectHacker** · October 28th 2008, 04:14 PM

Originally Posted by Mathstud28

$0\leq\left(\frac{3n}{4n+1}\right)^{2n}\leq\left(\f rac{3n}{4n}\right)^{2n}=\left(\frac{9}{16}\right)^ n$

A convergent geometric series

Yes this way is nicer.

**Hweengee** · October 29th 2008, 09:29 AM

very nice, thanks guys.

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