Infinite Series: Finding the Sum and General Term for Convergent Power Series

[sobek]

http://www4a.wolframalpha.com/Calculate/MSP/MSP167199e5bhg1gg5673i000048ed16i8cbf5iacg?MSPStoreType=image/gif&s=14&w=256&h=40

for all x in the interval of convergence of the given power series.

a) write the first 3 nonzero terms and the general term for an infinite series that represents http://www4a.wolframalpha.com/Calculate/MSP/MSP34199e5dhc325d3hd200005ag14201e835f640?MSPStoreType=image/gif&s=52&w=70&h=36

b) Find the sum of the series as determined in part a.

I have no idea how to do these two parts. I just don't know where to begin. Any help is appreciated thanks.

 

[rock.freak667]

Your general term is

\frac{(n+1)x^n}{3^{n+1}}

so your sum can be represented as

f(x) = \sum_{n=0} ^N \frac{(n+1)x^n}{3^{n+1}}

Now when you integrate within the summation sign, you just integrate what is being summed

e.g.

\int \sum_{n=0} ^N f(x) dx = \sum_{n=0} ^N \int f(x) dx

 

[sobek]

Okay so I can just take the integral of the general term and find the sum of the first 3 nonzero terms?

 

[rock.freak667]

Okay so I can just take the integral of the general term and find the sum of the first 3 nonzero terms?

Yes, then you can write out the first three terms and easily find a closed form for the sum.

 

[sobek]

Ok, I understand that, but what exactly are the limits of integration (0,1) there for?

 

[rock.freak667]

Ok, I understand that, but what exactly are the limits of integration (0,1) there for?

Because you will need to integrate between those limits

\sum_{n=0} ^N \int_0 ^1 f(x) dx

 

[sobek]

Agh! Ok I'm confused. If I'm just taking the integral of the general form, do I still need those limits? Or am I doing a Fundamental Theorem thing?

 

[rock.freak667]

Agh! Ok I'm confused. If I'm just taking the integral of the general form, do I still need those limits? Or am I doing a Fundamental Theorem thing?

yes you would still need those limits. It would be equivalent of doing the integral between those limits of every term.


Thus you just need to compute

\int_0 ^1 \frac{(n+1)x^n}{3^{n+1}}dx

 

[sobek]

Okay so how exactly would I integrate this? Seeing as there are two variables.

 

[rock.freak667]

Okay so how exactly would I integrate this? Seeing as there are two variables.

Treat 'n' as a constant.

 

[sobek]

Ok so for the integral I got http://www4a.wolframalpha.com/Calculate/MSP/MSP758199e8ha121cf0a4c00004ggc11fi1gf83a0h?MSPStoreType=image/gif&s=29&w=34&h=40

Is that right?

and from that is the sum: infinity?

 

[rock.freak667]

Ok so for the integral I got http://www4a.wolframalpha.com/Calculate/MSP/MSP758199e8ha121cf0a4c00004ggc11fi1gf83a0h?MSPStoreType=image/gif&s=29&w=34&h=40

Is that right?

and from that is the sum: infinity?

substitute the limits of the integral.


But now your sum will go from n=-1 to ∞

 

[sobek]

Okay with the limits of integration substituted I got http://www2.wolframalpha.com/Calculate/MSP/MSP344199e931ddgafc99d0000507h04h0277efcda?MSPStoreType=image/gif&s=9&w=34&h=37

and then i find the first 3 nozero terms from that right?

 

[Mathnerdmo]

Now, I can't see the original question (the link to wolframalpha is dead), but this doesn't sound right.

Can you post the problem again, sobek?

 

[sobek]

http://www2.wolframalpha.com/Calculate/MSP/MSP608199e96816fc40d8a000022aig2805fea6cad?MSPStoreType=image/gif&s=10&w=65&h=38

for all x in the interval of convergence of the given power series.

a) write the first 3 nonzero terms and the general term for an infinite series that represents http://www2.wolframalpha.com/Calculate/MSP/MSP179199e974h5cca8ib200004hf8af3ab6c4i576?MSPStoreType=image/gif&s=12&w=70&h=36

b) Find the sum of the series as determined in part a.

I'm still having trouble on part A

 

[Mathnerdmo]

Okay, I see it now.

Okay with the limits of integration substituted I got http://www2.wolframalpha.com/Calculate/MSP/MSP344199e931ddgafc99d0000507h04h0277efcda?MSPStoreType=image/gif&s=9&w=34&h=37

and then i find the first 3 nozero terms from that right?

Yes.

Assuming your original power series started at n=0, start from there.

Let n=0 in \frac 1{3^{n+1}} to find the first term. Then let n=1 to find your next, etc.

 

[sobek]

Okay so my math in getting https://www.physicsforums.com/latex_images/26/2602939-0.png was right?

and because it's a geometric series I can use a/1-r to find the sum right?

 

[Mathnerdmo]

Yes and yes.

 

[sobek]

Thanks to rocknerd667 and mathnerdmo for helping me with this problem