Class 25 - Ratio test + Power series (11/2/2017, Thursday)
Logistics
Exam 2- 11/13 Mon
- Check Syllabus on the course website. Sections 9.1 - 9.5 were dealt in Final Exams in the past, so you want to take note of that when you do problems from past exams.
WebHW
- 9.4 due 11/6 Mon
Lecture
A finer technique for alternating series. We now introduce another motivating exercise:
Exercise. Estimate s = \sum_{n=1}^{\infty}(-1)^{n}/n within the error of 0.1.
Error bound for alternating series. The key to above the exercise is to note the following inequality about the partial sums:
|s - s_{n}| \leq |s_{n+1} - s_{n}| = a_{n+1}.
Hence, we only need to take n = 9 to have a_{n+1} = a_{10} = 1/10 = 0.1. This means that our estimation is
s_{10} = 1 + 1/2 + \cdots + 1/10.
Example. Converge/diverge?
\sum_{n=1}^{\infty}\frac{1}{n!}.
This is a very easy question. However, I will introduce somewhat lengthier solution to introduce another technique. Write
a_{n} = 1/n!,
and consider
\frac{a_{n+1}}{a_{n}} = \frac{n!}{(n+1)!} = \frac{1}{n+1}.
We have \lim_{n \rightarrow \infty}a_{n+1}/a_{n} = 0. This means that if we pick N so large, then for any n \geq N, we have
0 \leq a_{n+1}/a_{n} \leq 1/2.
Therefore, we have
0 \leq a_{N+1} \leq (1/2)a_{N}
when n \geq N. Hence, for any m \geq 0, we have
0 \leq a_{N+m} \leq (1/2)a_{N+m-1} \leq (1/2)^{2}a_{N+m-2} \leq \cdots \leq (0.1)^{m}a_{N},
so
0 \leq \sum_{n=N}^{\infty}a_{n} = \sum_{m=0}^{\infty}a_{N+m} \leq \sum_{m=0}(1/2)^{m}a_{N} = 2a_{N} < \infty.
This implies that \sum_{n=N}^{\infty}a_{n} converges. The same argument (with some more work) shows the following:
Ratio test. Given any sequence (a_{n})_{n=1}^{\infty}
Power series. A power series is an "infinite extension" of a polynomial. It is useful because we can write something like:
e^{x} = 1 + x + x^{2}/2! + x^{3}/3! + \cdots.
As long as we know the right-hand side makes sense and term-by-term differentiable, the fundamental theorem of calculus gives us the equality.
Generally the power series around x = 0 looks like
f(x) = \sum_{n=0}^{\infty}a_{n}x^{n},
but we are only allowed to put x-values such that the sum converges. The ratio test tells us that if \lim_{n \rightarrow \infty}|a_{n+1}|/|a_{n}| = 1/R (allowing R = \infty in the case when the limit is 0), the ratio test tells us that R is the radius of (absolute) convergence of the power series.
Exercise. Figure out why!
Exercise*. Do #11 of Exam 3, Winter 2016.
Exercise. Estimate s = \sum_{n=1}^{\infty}(-1)^{n}/n within the error of 0.1.
Error bound for alternating series. The key to above the exercise is to note the following inequality about the partial sums:
|s - s_{n}| \leq |s_{n+1} - s_{n}| = a_{n+1}.
Hence, we only need to take n = 9 to have a_{n+1} = a_{10} = 1/10 = 0.1. This means that our estimation is
s_{10} = 1 + 1/2 + \cdots + 1/10.
Example. Converge/diverge?
\sum_{n=1}^{\infty}\frac{1}{n!}.
This is a very easy question. However, I will introduce somewhat lengthier solution to introduce another technique. Write
a_{n} = 1/n!,
and consider
\frac{a_{n+1}}{a_{n}} = \frac{n!}{(n+1)!} = \frac{1}{n+1}.
We have \lim_{n \rightarrow \infty}a_{n+1}/a_{n} = 0. This means that if we pick N so large, then for any n \geq N, we have
0 \leq a_{n+1}/a_{n} \leq 1/2.
Therefore, we have
0 \leq a_{N+1} \leq (1/2)a_{N}
when n \geq N. Hence, for any m \geq 0, we have
0 \leq a_{N+m} \leq (1/2)a_{N+m-1} \leq (1/2)^{2}a_{N+m-2} \leq \cdots \leq (0.1)^{m}a_{N},
so
0 \leq \sum_{n=N}^{\infty}a_{n} = \sum_{m=0}^{\infty}a_{N+m} \leq \sum_{m=0}(1/2)^{m}a_{N} = 2a_{N} < \infty.
This implies that \sum_{n=N}^{\infty}a_{n} converges. The same argument (with some more work) shows the following:
Ratio test. Given any sequence (a_{n})_{n=1}^{\infty}
- If \lim_{n \rightarrow \infty}|a_{n+1}|/|a_{n}| = c < 1, then \sum_{n=1}^{\infty}|a_{n}| converges.
- \lim_{n \rightarrow \infty}|a_{n+1}|/|a_{n}| = c > 1 then \sum_{n=1}^{\infty}|a_{n}| diverges.
Ratio 1 is useless. Consider \sum_{n=1}^{\infty}1/n and \sum_{n=1}^{\infty}(-1)^{n-1}/n.
Power series. A power series is an "infinite extension" of a polynomial. It is useful because we can write something like:
e^{x} = 1 + x + x^{2}/2! + x^{3}/3! + \cdots.
As long as we know the right-hand side makes sense and term-by-term differentiable, the fundamental theorem of calculus gives us the equality.
Generally the power series around x = 0 looks like
f(x) = \sum_{n=0}^{\infty}a_{n}x^{n},
but we are only allowed to put x-values such that the sum converges. The ratio test tells us that if \lim_{n \rightarrow \infty}|a_{n+1}|/|a_{n}| = 1/R (allowing R = \infty in the case when the limit is 0), the ratio test tells us that R is the radius of (absolute) convergence of the power series.
Exercise. Figure out why!
Exercise*. Do #11 of Exam 3, Winter 2016.
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