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