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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