1. ## comparison test

How would you work out these problems to find out if they converge or diverge?

Problem 1

n=1 to infinity 1/(square root (n^3 +2))

Problem 2

n=1 to infinity n+2^n/ n^2 2^n

Problem 3

n=1 to infinity 3^(n-1) + 1/ 3^n

2. Hello, TAG16!

Converge or diverge?

$\displaystyle (1)\;\;S \;=\;\sum^{\infty}_{n=1} \frac{1}{\sqrt{n^3+2}}$
We have: .$\displaystyle \frac{1}{\sqrt{n^3+2}} \:<\:\frac{1}{\sqrt{n^3}} \:=\:\frac{1}{n^{\frac{3}{2}}}$

$\displaystyle \text{Hence: }\;\sum^{\infty}_{n=1}\frac{1}{\sqrt{n^3+2}} \quad <\; \!\!\! \underbrace{\sum^{\infty}_{n=1}\frac{1}{n^{\frac{3 }{2}}}}_{\text{convergent }p\text{-series}}$

Therefore, $\displaystyle S$ converges.

$\displaystyle (2)\;\;S \;=\;\sum^{\infty}_{n=1} \frac{n+2^n}{n^2\!\cdot\!2^n}$

We have: .$\displaystyle \sum \frac{n+2^n}{n^2\!\cdot\!2^n} \;=\;\sum\left(\frac{n}{n^2\!\cdot\!2^n} + \frac{2^n}{n^2\!\cdot\!2^n}\right) \;=\;\underbrace{\sum\frac{1}{n\!\cdot\!2^n}}_A + \underbrace{\sum\frac{1}{n^2}}_B$

Series $\displaystyle A\!:\;\;\sum\frac{1}{n\!\cdot\!2^n} \:\leq\:\sum\frac{1}{2^n}$ . . . a convergent geometric series.
. . Hence, $\displaystyle A$ converges.

Series $\displaystyle B\!:\;\;\sum\frac{1}{n^2}$ . . . a convergent $\displaystyle p$-series.
. . Hence, $\displaystyle B$ converges.

The sum of two convergent series is convergent.

Therefore, $\displaystyle S$ converges.

$\displaystyle (3)\;\;S \;=\;\sum^{\infty}_{n=1}\frac{3^{n-1} + 1}{3^n}$

We have: .$\displaystyle \frac{3^{n-1}+1}{3^n} \;=\;\frac{3^{n-1}}{3^n} + \frac{1}{3^n} \;=\;\frac{1}{3} + \frac{1}{3^n} \;{\color{blue}\;>\;\frac{1}{3}}$

Hence: .$\displaystyle \sum\frac{3^{n-1}+1}{3^n} \;> \;\sum\frac{1}{3}$ . . . a divergent series

Therefore, $\displaystyle S$ diverges.