Find the sum of the following series

[verd]

Hi,

I'm having a bit of difficulty with the following two problems:

The first asks to find the sum of the following series:
\sum\limits_{n = 1}^\infty {\frac{2}{n^2+4n+3}}

I easily understand geometric series and how to find sums with those, but this seems a bit complicated. Factoring and pulling the 2 out doesn't seem to really help any. ...Any suggestions?


And this question:

Find a power series representation for the following:
g(x)=\arctan{\frac{x}{2}}

I understand the simple 1/4+x^2 kind of stuff, but this I don't even know how to begin.


Any suggestions??


Thanks!

 

[verd]

...Telescoping series for #1??

 

[Geekster]

\sum\limits_{n = 1}^\infty {\frac{2}{n^2+4n+3}}

I would suggest checking to see if partial fraction decomposition on the summand makes the problem easier to get at.

After you break it apart you should get something like

<br /> \sum\limits_{n = 1}^\infty {\frac{1}{n+1}} - \sum\limits_{n = 1}^\infty {\frac{1}{n+3}}<br />

Then notice that this telescoping series has only two terms that don't get canceled out. Add those two terms to get the sum.

 

[verd]

Wouldn't that work for testing convergence/divergence?

I need to find the sum... I wasn't aware that by partial fractions/integrals you could find sums... Is that possible? If so, what are the conditions?

 

[Geekster]

Find a power series representation for the following:
g(x)=\arctan{\frac{x}{2}}

What's the derivative of arctan(x)?

Can you find a power series representation for that? Can you integrate term by term?

 

[Geekster]

Wouldn't that work for testing convergence/divergence?

I need to find the sum... I wasn't aware that by partial fractions/integrals you could find sums... Is that possible? If so, what are the conditions?

That works fine for the sum. The partial faction decomp isn't just for integrals. Try adding the two sums together and see that you get back your original summand. Once you're convinced they are in fact the same, then notice that every n+2 for the first summand is the additive inverse of the nth term of the second summand. You should notice that only two terms are not canceled out in this manner. Add those two terms to get the sum.

 

[verd]

Woo, I figured it out right before you replied! Haha, thanks, you verified that what I was doing was correct! Thanks!

Anyone have any ideas on that second one?

 

[Geekster]

Anyone have any ideas on that second one?

Look three post up from this one...

 

[verd]

Can someone tell me if this is correct for the second problem posted?

\sum\limits_{n = 1}^\infty {\frac{(-1)^{n}x^{2n-1}}{(2n-1)4^{n-1}}}

 

[Curious3141]

Can someone tell me if this is correct for the second problem posted?

\sum\limits_{n = 1}^\infty {\frac{(-1)^{n}x^{2n-1}}{(2n-1)4^{n-1}}}

The easiest way (non-rigorous, but fast) to verify your answer is to work out around five to six terms with some "not so nice" value for x (like 0.23) and confirm that the sum is tending to the correct value for arctan(0.23/2). Quite fast with a scientific calc and the Memory+ function.