# [Calculus] Infinite series

1. Jan 6, 2010

### WannaBe22

1. The problem statement, all variables and given/known data

I'm kind of new around here...Hope this forum will help me be a better future scientist :)
I need some serious help in the attached questions...I need to determine whether the series in the picture converge, absolutely converge ot diverge...
I realy need your guidance and help....

Thanks everyone

2. Relevant equations
3. The attempt at a solution
I've no clue aboue 1&3...
In 2-I know that cos(n*pi)*ln([n+1]/n ) converges by leibnitz test...But what about the first element in the sum? [sqrt(n)sin(n)tan(1/n^2)] ...

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2. Jan 6, 2010

### Dick

Here are some hints. For 3) try simplifying 1/(sqrt(n)-1)-1/(sqrt(n)+1). For the first part of 2) try comparing it to a power series. You should know if x is very small x~tan(x). For 1) try expanding sin(1/n) in a power series, and then looking at the results.

3. Jan 6, 2010

### penguin007

The first one converges absolutly(and therefore converges) because each term is equivalent to 1/n^3 (then compare with Riemann's series);
For the last one, you can split the sum in order to eliminate "many" terms.

Last edited: Jan 6, 2010
4. Jan 6, 2010

### WannaBe22

About 1- You've said each term in the series is equivalent to 1/n^3, why is that? Have you used Taylor's Series and if you did, how did you use it?

About 2- So we know that when lim_x->infinity_[tan(1/n^2)/(1/n^2)]=1...But how can this help us and to which power series can I compare it to?

I've tried this:
Each two close elements can be written as follow:
1/(sqrt(n)-1) - 1/(sqrt(n)+1) = 2/n... We know the series Sigma_1/n diverges and so does Sigma_2/n ofcourse...But how can we compare the relevant series to 2/n if each two elements can be compared and not each one...? I mean...We can simplify the sum only by simplifying each 2 close elements, but it's not allowed (Because if we take the series 1-1+1-1+1-1... We can Write it as (1-1)+(1-1)+... and it converges to 0 or to write it as 1+(-1+1)+(-1+1)+... and it converges to 1 etc... ) ...So how can we make use of our "discovery"?

Thanks to you both and I hope you'll be able to continue guiding me...

5. Jan 6, 2010

### Dick

About 2. According to your limit, you can say for example tan(1/n^2)<=2/n^2 for n sufficiently large. About 3, good point! But remember the sum of the infinite series is the limit of the partial sums. You've shown the even partial sums will diverge since it's the sum of 2/(n-1). Then the odd partial sums will also diverge, since the terms tend to zero, right? In general if you can regroup (NOT reorder) a series and make it divergent, the original series was divergent. You'll have to ask penguin007 about the 1/n^3.

6. Jan 7, 2010

### WannaBe22

Thanks a lot to both of you! I don't need any further help ::)