# Math Help - Infinite Series

1. ## Infinite Series

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.

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

3. 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
$

4. 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.

5. 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

6. 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.

7. very nice, thanks guys.