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The McLen series for a function ( f ) is given by, and they give it in Sigma notation, and then they expand it out for us. It converges to ( f(x) ) for the absolute value of ( x ) being less than ( R ), where ( R ) is the radius of convergence of the McLen series.

Part A: Use the ratio test to find ( R ). So first of all, if terms like McLen series and radius of convergence, or even convergence or ratio test seem foreign to you or you have some foggy memories of it, you might want to review all of those concepts on KH Academy. We actually have multiple videos and exercises on each of these concepts on KH Academy.

But if you kind of know what it is, I will give you a little bit of a reminder for the ratio test. The ratio test tells us if we have an infinite series, so we go from ( n = 1 ) to infinity, and each term is ( a_n ). The ratio test says, all right, let's consider the ratio between successive terms. So we could say the ratio of ( a_{n+1} ) over ( a_n ), and in particular, we want to focus on the absolute value of this ratio.

By itself, it might not be a constant like we would see in a geometric series; it actually might be a function of ( n ) itself. So we want to see the behavior of this ratio as ( n ) gets really, really, really large, as we're kind of, you know, adding those terms as we're getting close to infinity.

So we'd want to take the limit as ( n ) approaches infinity here. If this limit exists, let's say it's ( L ). If ( L < 1 ), then the series converges. If ( L > 1 ), it diverges. If it's equal to 1, it's inconclusive.

What we want to do here is figure out the absolute value of the ratio, take the limit, and then see for what ( x ) values that limit will be less than one. So let's do that.

Let's first think about this ratio. So ( a_{n+1} ) over ( a_n ) is going to be equal to... So if we put ( n + 1 ) into this expression here, we're going to have... So let me make this clear.

So I'm going to do ( a_{n+1} ) up here. So we're going to have ( -3^{n+1} ) instead of ( -3^n ), and then times ( x^{n+1} ) over ( (n+1) ). So that's ( a_{n+1} ) there, and ( a_n ) is just ( -3^n \cdot x^n ) over ( n ).

What does this right over here simplify to? This is going to be equal to... Well, we could just say this divided by that; it's the same thing as multiplying by the reciprocal of all of this stuff down here. So it's ( -3^n \cdot x^{n+1} ) over ( (n+1) ) times the reciprocal of this business, so times ( \frac{n}{-3^n \cdot x^n} ).

Can we simplify this? Well, we can divide both the numerator and the denominator here by ( x^n ). So this divided by ( x^n ) is just ( 1 ), and this divided by ( x^n ) is going to be ( x ) or ( x^1 ). We can divide the numerator and the denominator by ( -3^{n-1} ). Well, this is just going to be ( 1 ), and if you divide ( -3^n ) by ( -3^{n-1} ), well that's just going to be ( -3^1 ).

So this is all going to be ( \frac{3x}{n + 1} ). Now let's think about what the limit of the absolute value of this as ( n ) approaches infinity is. So the limit as ( n ) approaches infinity of the absolute value of ( \frac{-3x}{n + 1} ).

Now some of you might recognize if we focus on ( n ), we have the same degree up here, same degree down here; both are ( n^1 ). So ( \frac{n}{n + 1} ) is going to approach ( 1 ), and so you might say, okay well, this is going to be the absolute value of ( -3x ).

But if you want to make that a little bit clearer, this is equal to the limit as ( n ) approaches infinity of... and I'll write it this way. Let me write it so I could write the absolute value of ( -3x ) or the absolute value of ( -3x ) is the same thing as ( 3 \times \text{absolute value of } x \times \text{absolute value of } \ldots ).

If I divide the numerator here by ( n ), I would get ( 1 ), and if I divide the denominator by ( n ), I can do... as long as I multiply or divide the denominator and the numerator by the same thing, I'm not changing the value. So if I divide both of them by ( n ) in the numerator, I just get ( 1 ). In the denominator ( n/n ) is ( 1 ); ( 1/n ) is ( +1/n ).

This might be a little bit clearer that, okay, as ( n ) approaches infinity... well, we don't know; this doesn't deal with ( n ), but this over here ( 1/n ) is going to approach ( 0 ). And so this whole thing is going to approach ( 1 ), and so the limit is going to be ( 3 \times \text{absolute value of } x ).

So remember, this series converges if this limit is less than one. So it converges if ( 3 \times \text{absolute value of } x < 1 ). We could say the absolute value of ( x ) divided both sides by ( 3 ) is less than ( \frac{1}{3} ).

So we have just found our radius of convergence ( R ). We could say ( R ) is equal to ( \frac{1}{3} ). This McLen series is going to converge as long as the absolute value of ( x ) is less than ( \frac{1}{3} ), or we could say our radius of convergence is equal to... our radius of convergence is equal to ( \frac{1}{3} ).

So there you go.