Regarding Remainder of an Alternating Series

Alternating series test (Wikipedia.org)

Definition of an alternating series:
A series of the form $$\sum_{k=0}^\infty (-1)^k a_k$$
where all the $a_k$ are non-negative. The emphasis here is on the non-negative.

We know the absolute value of the remainder $R_n=\sum_{k=n+1}^\infty (-1)^k a_k$ for an alternating series will be less than the next term in the series, $a_{n+1}$.

So what's the point of all this? Well, some power series might look like alternating series, but they'll only be alternating series for certain values of $x$. Thus, you can only estimate the remainder using the above estimate for values of $x$ for which the power series turns into an alternating series.

[Remark: It sort of follows from the above, that a power series that doesn't look like an alternating series, might be an alternating series for certain values of $x$]

Answering a student's email questions

For question 6, you should set $x=0$ and solve for $t$. You'll get two values of $t$, but only the one that gives you $y\geq 3$ will work. Then set $y=3$ and solve for $t$. Again you'll get two values of $t$, but only one gives you $x\geq 0$.

For question 7, the furthest distance from the origin is when $r$ is greatest. Thus we are trying to maximize $r$. Then from one variable calculus, just take the derivative with respect to $\theta$ to find critical points. $\frac{dr}{d\theta}=0$. There's a geometric interpretation to this, but it's much easier to explain through pictures than through words. If requested, I'll explain this.

Question 8 is asking for when $r=0$. No derivatives required. Let me know if I'm wrong on this.

For question 10, your mistake might have been that the integral should go from $0$ to $\pi$ and not from $0$ to $2\pi$.

Taylor Remainder Term

On page 756 (of Stewart's Single Variable Calculus Early Transcendentals), the margin has the following information:
If $f^{(n+1)}$ is continuous on an interval $I$ and $x\in I$, then $$R_n(x)=\frac{1}{n!}\int_a^x(x-t)^n f^{(n+1)}(t)dt$$. This is called the integral form of the remainder term.

Lagrange's form of the remainder term:
$$R_n(x)=\frac{f^{n+1}(z)}{(n+1)!}(x-a)^{n+1})$$
for some $z$ between $x$ and $a$.

Note that Taylor's Inequality follows from Lagrange's form of the remainder term.

Let's do a similar problem to #17 on this the 2010 final exam.

$f(x)=3^x$ and $a=5$.
$f^\prime(x)=\ln(3) 3^x$
$f^{\prime\prime}(x)=\ln(3)^2 3^x$
$f^{(3)}(x)=\ln(3)^3 3^x$
Then $f$ has Taylor series $$\sum_{n=0}^\infty \frac{\ln(3)^n 3^5}{n!}(x-5)^n$$
It's remainder $R_n(x)$ will be $$R_n(x)=\frac{1}{n!}\int_5^x(x-t)^n \ln(3)^{n+1} 3^t dt$$
Note: Using this formula, I would answer the following for part (c): $$R_2(2.1)=\frac{1}{2!}\int_2^{2.1} (2.1-t)^2 \ln(2)^3 2^t dt$$
To answer part (d), just integrate the above by parts and use all the information he gave you to simplify the answer. You might get a different answer since he gave an answer I don't fully understand to part (c). See Wikipedia: Taylor's theorem > Explicit formulae for the remainder

part of page 762

$$\frac{1}{1-x}=\sum_{n=0}^\infty x^n$$
$$e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$$
$$\sin x=\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{(2n+1)!}$$
$$\cos x=\sum_{n=0}^\infty (-1)^n\frac{x^{2n}}{(2n)!}$$
Note: $\sin 0=0$ and $\cos 0=1$
$$\tan^{-1} x=\sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{2n+1}$$
$$\ln(1+x)=\sum_{n=1}^\infty (-1)^{n-1}\frac{x^n}{n}$$
$$(1+x)^{k}=\sum_{n=0}^{\infty}\left(\begin{array}{c}
k\\
n\end{array}\right)x^{n}$$
$$\left(\begin{array}{c}
k\\
n\end{array}\right)=\frac{k(k-1)(k-2)\cdots(k-n+1)}{n!}$$