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- Ch. 3
- 6. ...
- 7. If $\sum a_n$ converges, $\sum\frac{\sqrt{a_n}{n}$ converges, for consider that for any n either $\frac{\sqrt{a_n}{n}$ is greater, equal to or less than $a_n$; if $\frac{\sqrt{a_n}{n} > a_n$ then $a_n < n^{-2}$ and $\frac{\sqrt{a_n}{n} < n^{-2}$. So since any given term in $\{\frac{\sqrt{a_n}{n}\}$ is less than the corresponding term either in $\{a_n\}$ or in $\{n^{-2}\}$, $\sum\frac{\sqrt{a_n}{n}$ must be less than the sum of these latter two sequences, which is finite. So $\sum\frac{\sqrt{a_n}{n}$ converges.
- 8. If $\sum a_n$ converges, and $\{b_n\}$ is monotonic and bounded, $\sum a_nb_n$ converges; for there is some M so $|b_n| \leq M$ for all n and $\sum a_nb_n \leq \sum Ma_n = M\sum a_n$.
- 9.
- 10. Suppose that the coefficients of ...
- 11.
- 16.
- 17.
- Ch. 4.
- 1.
- 2.
- 3. Let $Z(f)$ be the set of all points p in a metric space X so f(p)=0 for some real-valued continuous function f. Suppose Z is not closed. Then some q exists in X which is a limit point of Z and not in Z, and f(q) is $\alpha$. Now for any $\epsilon$ there is a $\delta$ so if $ d_x(x,q) < \delta$, $|\alpha| < \epsilon$ (if x is chosen as one of the points in Z which must exist within any $\delta$-neighborhood of q). Obviously some $\epsilon$ can be chosen so $|\alpha| > \epsilon$. But then f is not continuous at all points. The supposition must be false and Z is closed.
- 4.f and g are continuous mappings
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