Query 13

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GOOD STUDENT SOLUTION WITH COMMENT: For large values of n, this series is similar to 1 / 3^n. As n approaches infinity, 1 / 3^n approaches 0. A larger number in the denominator means that the value of the function will be smaller. So, (1 / 3^n) > (1 / (3^n + 1)). We know that since 1 / 3^n converges, so does 1 / (3^n + 1).

COMMENT: We know that 1 / (3^n) converges by the ratio test: The limit of a(n+1) / a(n) is (1 / 3^(n+1) ) / (1 / 3^n) = 1/3, so the series converges.

We can also determine this from an integral test. The integral of b^x, from x = 0 to infinity, converges whenever b is less than 1 (antiderivative is 1 / ln(b) * b^x; as x -> infinity the expression b^x approaches zero, as long as b < 1).

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*&*& The ratio test takes the limit as n -> infinity of a(n+1) / a(n). If the limit is less than 1 then the series converges in much the same way as a geometric series sum(r^n), with r equal to the limiting ratio.

In this case a(n+1) = 1 / (2n + 2)! and a(n) = 1 / (2n) ! so

a(n+1) / a(n)

= 1 / (2n+2) ! / [ 1 / (2n) ! ]

= (2n) ! / (2n + 2) !

= [ 2n * (2n-1) * (2n-2) * (2n - 3) * ... * 1 ] / [ (2n + 2) * (2n + 1) * (2n) * (2n - 1) * ... * 1 ]

= 1 / [ (2n+2) ( 2n+1) ].

As n -> infinity this result approaches zero. Thus the series converges for all values of n, and the radius of convergence is infinite. *&*&

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We have:

an = 10^(-n)

and

an+1 = 10^(-(n+1))

So, since 0 < an+1 < an, this series converges.

*&*& 0 < an+1 < an is not the appropriate test. For example if a(n) = (-1)^n * (.5 + 10^-n) we have the series .5 - .51 + .501 - .5001 etc. and the partial sums jump back and forth by about .5 units and never approach a limit.

What you have is an alternating series where | a(n) | -> 0. This is the criterion for convergence of an alternating series. *&*&

This is an alternating series with | a(n) | = .1^n, for n = 0, 1, 2, ... .

Thus limit{n->infinity}(a(n)) = 0.

An alternating series for which | a(n) | -> 0 is convergent.

sum(1/n^.999) diverges and sum(1 / n^1.001) converges, but doing partial sums on your calculator will never reveal this. The calculator is very limited in determining convergence or divergence.

However there is a pattern to the partial sums, which are 1, .9, .91, .909, .9091, .90909, ... . It's easy enough to show that the pattern continues, so the convergent value is .9090909... .

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** The general term is the coefficient of x^n.

In this case you would have the factors of x^2, x^3 etc. go down to p-1, p-2 etc.; the last factor is p minus one greater than the exponent of x. So the factor of x^n would go down to p - n + 1.

This factor would therefore be p ( p - 1) ( p - 2) ... ( p - n + 1).

This expression can be written as p ! / (p-n) !. All the terms of p ! after (p - n + 1) will divide out, leaving you the desired expression p ( p - 1) ( p - 2) ... ( p - n + 1).

The nth term is therefore (p ! / (p - n) !) / n! * x^n, of the form a(n) x^2 with a(n) = p ! / (n ! * (p - n) ! )

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To find the radius of convergence you first find the limit of the ratio | a(n+1) / a(n) | as n -> infinity. The radius of convergence is the reciprocal of this limit.

a(n+1) / a(n) = ( n / (2n+1) ) / (n+1 / (2(n+1) + 1) ) = (2n + 3) / (2n + 1) * n / (n+1).

(2n + 3) / (2n + 1) = ( 1 + 3 / (2n) ) / (1 + 1 / (2n) ), obtained by dividing both numerator and denominator by 2n. In this form we see that as n -> infinity, this expression approaches ( 1 + 0) / ( 1 + 0) = 1.

Similarly n / (n+1) = 1 / (1 + 1/n), which also approaches 1.

Thus (2n + 3) / (2n + 1) * n / (n+1) approaches 1 * 1 = 1, and the limit of a(n+1) / a(n) is therefore 1.

The radius of convergence is the reciprocal of this ratio, which is 1.

@&

It's much more rigorous to say that 3 / (2 n) approaches 0 than to say that the 1 and the 3 are inconsequential. They are, and it's pretty clear why, so you can get away with that here. But the rigorous approach is just as quick as stating that a couple of numbers are inconsequential, and stands on a much firmer base.

Also there are places you're going to need that technique, where it's much less clear what's consequential and what isn't.

*@

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*&*& As seen in 9.4.6 we have

a(n) = p ! / (n ! * (p - n) ! ) so

a(n+1) = p ! / [ (n+1) ! * ( p - (n+1) ) ! ] and

a(n+1) / a(n) = { p ! / [ (n+1) ! * ( p - (n+1) ) ! ] } / {p ! / (n ! * (p - n) ! ) }

= (n ! * ( p - n) !) / {(n+1)! * (p - n - 1) ! }

= (p - n) / (n + 1).

This expression can be written as

(p / n - 1) / (1/n + 1). As n -> infinity both p/n and 1/n approach zero so our limit is -1 / 1 = -1.

Thus the limiting value of | a(n+1) / a(n) | is 1 and the radius of convergence is 1/1 = 1. *&*&

@&

You did a good job.

See my note on the preceding question.

It's a little less obvious here than p is inconsequential, and the rigorous proof is much more airtight.

*@

&#This looks good. See my notes. Let me know if you have any questions. &#