limits

**DontKnoMaff** · October 13th 2010, 06:09 PM

PROBLEM STATEMENT:

(a)Suppose that { $a_n$ } and { $b_n$ } are sequences of nonnegative numbers and let $c_n = max(a_n,b_n)$ . Show that $\sum c_n$ converges iff both $\sum a_n$ and $\sum b_n$ converge. Hint: $c_n \leq a_n+b_n$ .

(b)Prove that if $\sum a_n$ converges, where $a_n \geq 0$ , then $\sum \frac{\sqrt{a_n}}{n}$ converges, too. Hint: We already know that $\sum \frac{1}{n^2}$ converges.

MY WORK:

(a) We use the comparison test here which says that if $|a_n| \leq c_n$ for $n \geq N_0$ , where $N_0$ is some fixed integer, and if $\sum c_n$ converges, then $\sum a_n$ converges.
Supposing that $\sum c_n$ converges, then the larger of $a_n$ and $b_n$ converges. Suppose $a_n \geq b_n$ . Since $\sum c_n$ converges, and $c_n=a_n$ , $\sum a_n$ converges. Now by the comparison test, $b_n$ converges. The same is true if $b_n \geq a_n$ .
Now, supposing that both $\sum a_n$ and $\sum b_n$ converge, since $c_n \leq a_n$ and $c_n \leq b_n$ , $\sum c_n$ must converge by the comparison test.

(b)Here, we use the Schwarz inequality which states that if $a_1,...,a_n$ and $b_1,...,b_n$ are complex numbers, then $|\sum^n_{j=1} a_j\overline{b_j}|^2 \leq \sum^n_{j=1} |a_j|^2\sum^n_{j=1} |b_j|^2$ .
From here, we directly get $\sum \frac{\sqrt{a_n}}{n} \leq \sum a_n \sum \frac{1}{n^2}$ . and since both $\sum a_n$ converges, and $\sum \frac{1}{n}$ converges, then $\sum \frac{\sqrt{a_n}}{n}$ converges.

**raphw** · October 13th 2010, 06:21 PM

a) Rather prove it differently, your proof seems a little odd.

You know that $\sum a_n$ and $\sum b_n$ converges, so $\sum a_n + b_n$ converges.

Since $c_n = \max\{a_n,b_n\}$ you know that $c_n \le a_n + b_n$ . Thus $\sum c_n \le \sum a_n+ b_n$ and by the comparison test the first series must converge.

b) Is perfectly fine.

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