Series and positive sequence

Question

Suppose $b_n$ is positive sequence of real numbers and $\sum_{n=1}^∞b_n$ converges.

Show that $\sum_{n=1}^∞\sqrt{7b_n^2+9b_n^3}$ converge and $\sum_{n=1}^∞\frac{sin(b_n)}{b_n}$ diverge.

My idea for the first is to find a bigger sequence that convgerge so that the smaller also converge. For the second i have no idea how to proceed.

Answer 1

Since $\sum b_n$ converges then $b_n \to 0, n \to \infty$ hence $\frac{sin(b_n)}{b_n} \to 1 \neq 0, n \to \infty$ hence $\sum \frac{sin(b_n)}{b_n}$ diverges

Answer 2

Considering $a_n$ = $sqrt( 7(b_n)^2 + 9(b_n)^3)$ we know that if $a_n / b_n$ converges to a l > 0 and the series of $b_n$ converges then the series of $a_n$ converges. All you have to do is calculate the limit of $a_n / b_n$

Answer 3

Not that the series $$\sum_{n=1}^∞b_n$$ converges implies that {$b_n$} converges to $0$.

Thus for large enough $n$, we have $$ \sqrt {7b_n^2+9b_n^3}\le \sqrt {7b_n^2+9b_n^2} = 4b_n$$ which implies that $$\sum_{n=1}^∞\sqrt {7b_n^2+9b_n^3} $$ converges.

For $$\sum_{n=1}^∞\frac{sin(b_n)}{b_n}$$ use the divergence test.

$$\frac{sin(b_n)}{b_n}$$ approaches $1$ instead of $0$.

Thus $$\sum_{n=1}^∞\frac{sin(b_n)}{b_n}$$ diverges.

Answer 4

Since $\sum_{n=1}^∞b_n$ converges, then for $n\to\infty$ it holds that $b_n\to0$.

• Using the Limit Comparison Test: $$ b_n\to0\Rightarrow \frac{\sqrt{7b_n^2 +9b_n^3}}{b_n}= \left\lvert b_n\right\rvert\frac{\sqrt{7 +9b_n}}{b_n}= b_n\frac{\sqrt{7 +9b_n}}{b_n}= \sqrt{7 +9b_n}\to 7 > 0 $$ Therefore, since $\sum_{n=1}^∞b_n$ converges, $\sum_{n=1}^∞\sqrt{7b_n^2+9b_n^3}$ converges as well.
  
  
• Using the Limit Test: $$ x\to0\Rightarrow\frac{sin(x)}{x}= \frac{(sin(x))'}{(x)'}=cos(x)\to 1\ \ \ \Leftrightarrow\ \ \ b_n\to0\Rightarrow\frac{sin(b_n)}{b_n}\to 1\ne0 $$ Therefore, $\sum_{n=1}^∞\frac{sin(b_n)}{b_n}$ diverges.