Radius of convergence question

[Johnny5]

Can someone show me how to compute the radius of convergence of the following power series?

 

[math] \sum_{n=0}^{n=\infty} \frac{nx^n}{2^{n+1}} [/math]

 

There are different ways to do it.

 

Regards

The first term is zero, so

 

[math] \sum_{n=1}^{n=\infty} \frac{nx^n}{2^{n+1}} = \sum_{n=1}^{n=\infty} \frac{n(n-1)!x^n}{(n-1)!22^{n}} = \frac{1}{2}\sum_{n=1}^{n=\infty} x^n \prod_{k=1}^{k=n} \frac{k}{2(k-1)}

[/math]

 

So

 

[math] f(k) = \frac{k}{2(k-1)} [/math]

 

In the sum, n goes to infinity, hence k goes to infinity.

 

The product from k=1 to n, of f(K), gives the nth coefficient of the power series.

 

So in the limit we have:

 

[math] \lim_{k \to \infty} f(k) = \lim_{k \to \infty} \frac{k}{2(k-1)} [/math]

 

And we can use L'Hopital's rule here, so:

 

[math] \lim_{k \to \infty} \frac{k}{2(k-1)} = \frac{1}{2} [/math]

 

Actually, let me put in x, so that we are looking at f(x,k), instead of just f(k).

 

[math]\sum_{n=1}^{n=\infty} \frac{nx^n}{2^{n+1}} = \sum_{n=1}^{n=\infty} \frac{n(n-1)!x^n}{(n-1)!22^{n}} = \frac{1}{2}\sum_{n=1}^{n=\infty} \prod_{k=1}^{k=n} \frac{xk}{2(k-1)}

 

[/math]

 

Hence:

 

[math] f(x,k) = \frac{xk}{2(k-1)} [/math]

 

Actually let's look at the product:

 

 

[math] f(n,x,k) = \prod_{k=1}^{k=n} \frac{xk}{2(k-1)} [/math]

 

The formula above, is the n-th term of the series.

 

Now, consider the limit as n approaches infinity:

 

[math] f(n,x,k) = \prod_{k=1}^{k=\infty} \frac{xk}{2(k-1)} [/math]

 

This is now the limit as k approaches infinity of f(x,k).

 

[math] \lim_{k \to \infty} f(x,k) = \lim_{k \to \infty}\frac{xk}{2(k-1)} =\frac {x}{2}

[/math]

 

for a fixed x, which does not depend upon k.

 

So as long as x<2, the nth term of the series is less than 1, as n approaches infinity.

 

I am considering series whose terms are positive right now.

 

The limit as k approaches infinity of f(k), is one over the radius of convergence of the series. Right now, I am looking for a proof of this.

 

Consider the harmonic series:

 

[math] H_n = \sum_{n=1}^{n=\infty} \frac{1}{n} = 1+1/2+1/3+1/4+...+1/n+... [/math]

 

Clearly, the nth term goes to zero, as n approaches infinity, but that does not mean that the series converges.

 

What has to be done, is to consider the sequence of partial sums, and ask whether or not that goes to zero, as n tends to infinity.

 

When n=2, we have:

 

1+1/2=2/2+1/2=3/2

 

when n=3, we have

 

3/2+1/3=9/6+2/6=11/6

 

when n=4 we have

 

11/6+1/4 = 44/24+6/24=50/24=25/12

 

when n=5 we have:

 

25/12+1/5=125/60+12/60=137/60

 

Thus, the first few terms of the sequence of partial sums is given by;

 

(1,3/2,11/6,25/12,137/60,...)

 

Let us list the first few terms, with each term having a common denominator:

 

(60/60,90/60,110/60,125/60,137/60,...)

 

Here is some stuff on limits of sequences:

 

limit of a sequence

 

First find f(k).

 

[math] H_n = \sum_{n=1}^{n=\infty} \frac{1}{n} = 1+ \sum_{n=2}^{n=\infty} \frac{(n-1)!}{n(n-1)!} = 1+ \sum_{n=2}^{n=\infty} \frac{(n-1)!}{n!}[/math]

 

[math] 1+ \sum_{n=2}^{n=\infty} \frac{(n-1)!}{n!} = 1+ \sum_{n=2}^{n=\infty} \prod_{k=2}^{k=n} \frac{k-1}{k} [/math]

 

Hence:

[math] f(k) = \frac{k-1}{k} [/math]

 

In the limit as k goes to infinity, f(k) approaches 1.

 

Here is Wolfram on the concept of limit.

 

Wolfram on limit

 

Now, the series H_n converges if and only if the sequence of partial sums converges.

 

Let me branch again in this discussion, to consider whether or not the following series converges (because I know something about the partial sums of this series, is particularly simple):

 

[math] \sum_{n=1}^{n=\infty} \frac{1}{n(n+1)} = 1/(1*2) + 1/(2*3) +1/(3*4) +... [/math]

 

 

Now, we can use partial fractions to rewrite the summand.

 

Now, I know of two ways to do this, the coefficient method, and the Heaviside cover up method.

 

I will use the coefficient method.

 

[math] \frac{1}{n(n+1)} = \frac{A}{n}+\frac{B}{n+1} [/math]

 

the goal is to now solve for A,B.

 

[math] \frac{1}{n(n+1)} = \frac{A}{n}+\frac{B}{n+1} = \frac{A(n+1) + nB}{n(n+1)} [/math]

 

From which it follows that:

 

[math] A(n+1) + nB = 1 [/math]

 

Hence:

 

[math] An+A + nB = 1 [/math]

 

Hence:

 

[math] (A+B)n+A = 1 [/math]

 

Hence A=1, and A+B=0, from which it follows that B=-1.

 

Therefore:

 

[math] \frac{1}{n(n+1)} = \frac{1}{n}-\frac{1}{n+1} [/math]

 

So we can now rewrite the orginal series, with an equivalent summand, in a different form:

 

[math] \sum_{n=1}^{n=\infty} \frac{1}{n(n+1)} = \sum_{n=1}^{n=\infty}\frac{1}{n}-\frac{1}{n+1} [/math]

 

Now, here is the formula for the NTH partial sum:

 

[math] \sum_{n=1}^{n=N}\frac{1}{n}-\frac{1}{n+1}= (1-1/2)+(1/2-1/3)+...+(\frac{1}{N}-\frac{1}{N+1}) [/math]

 

The series above is an example of a telescoping series. The intermediate terms cancel each other out.

 

So, we have a simple formula for the NTH partial sum, which is:

 

[math] 1-\frac{1}{N+1} [/math]

 

So this is the formula for the sequence of partial sums:

 

[math] S(N) = 1-\frac{1}{N+1} [/math]

 

And in the limit as N approaches infinity, S(N) approaches 1. Thus, the sequence of partial sums converges, thus the series converges.

 

So back to the harmonic series H_n.

 

The harmonic series converges if and only if its sequence of partial sums converges, so if the sequence of partial sums diverges then the harmonic series diverges.

 

[math]H_n = \sum_{n=1}^{n=\infty} \frac{1}{n} = 1+ \sum_{n=2}^{n=\infty} \frac{(n-1)!}{n!} = 1+ \sum_{n=2}^{n=\infty} \prod_{k=2}^{k=n} \frac{k-1}{k}

[/math]

 

The next step is to focus on the sequence of partial sums, as in the previous example.

[uncool]

One way to do this: ratio test.

Take the absolute value of the ratio of each term to the one before it. Take the limit of this as n goes to infinity. If the limit is less that 1, then the series converges. If it is greater, it diverges. If it hits exactly, then you have to figure out what happens.

-Uncool-

[Johnny5]

One way to do this: ratio test.

Take the absolute value of the ratio of each term to the one before it. Take the limit of this as n goes to infinity. If the limit is less that 1' date=' then the series converges. If it is greater, it diverges. If it hits exactly, then you have to figure out what happens.

-Uncool-[/quote']

 

I was thinking of that as a next move.

 

Let me ask you something.

 

When you are faced with figuring out whether or not a series converges,

 

what is the first mental step of your approach to solution?

 

The reason I ask, is precisely because there are so many different methods. There is the integral test, the comparison test, the ratio test, the root test, and quite a few others which I don't recall off the top of my head.

 

Does the particular series in question dictate your approach, or do you have a one approach solves all method.

 

Regards