1. ## Series

Is this series absolutely convergent

2/5 + (2*6)/(5*8) + (2*6*10)/(5*8*11) + (2*6*10*14)/(5*8*11*14) + ...

2. Originally Posted by amiv4
Is this series absolutely convergent

2/5 + (2*6)/(5*8) + (2*6*10)/(5*8*11) + (2*6*10*14)/(5*8*11*14) + ...
$\displaystyle \sum_{n=0}^{\infty}\frac{2\cdot{6}\cdot{10}\cdots{ 4n+2}}{5\cdot{8}\cdot{11}\cdot{3n+5}}=\sum_{n=0}^{ \infty}\frac{\prod_{k=0}^{n}(4k+2)}{\prod_{k=0}^{n }(3k+5)}$

Using the Ratio Test givse

\displaystyle \begin{aligned}\lim_{n\to\infty}\left|\frac{\prod_ {k=0}^{n+1}(4k+2)}{\prod_{k=0}^{n+1}(3k+5)}\cdot\f rac{\prod_{k=0}^{n}(3k+5)}{\prod_{k=0}^{n}(4k+2)}\ right|&=\lim_{n\to\infty}\left|\frac{4(n+1)+2}{3(n +1)+5}\right|\\ &=\frac{4}{3}>1 \end{aligned}

3. shouldnt it be 4k +2 in the first part instead of 2k+4. and i think that messes everything else up

4. Originally Posted by amiv4
shouldnt it be 4k +2 in the first part instead of 2k+4. and i think that messes everything else up
Yes, that was a typo. The solution still holds.

5. k I'm really not seeing how when u multiply those two together you get that. Cuz i did the ratio test myself and i dont see how those multiply together to get that. Can you explain

6. Let's write it explicitly.

Let: $\displaystyle a_{n} = \frac{2\cdot{6}\cdot{10}\cdots{4n+2}}{5\cdot{8}\cd ot{11}\cdots{3n+5}}$

Using the ratio test,
\displaystyle \begin{aligned} \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_{n}}\right| & = \lim_{n \to \infty} \left| \frac{2\cdot{6}\cdot{10}\cdots{4(n+1)+2}}{5\cdot{8 }\cdot{11}\cdots{3(n+1)+5}} \ \cdot \ \frac{5\cdot{8}\cdot{11}\cdots{3n+5}}{2\cdot{6}\cd ot{10}\cdots{4n+2}} \right| \end{aligned} (simply what $\displaystyle \frac{a_{n+1}}{a_{n}}$ is)

$\displaystyle {\color{white}\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_{n}}\right|} = \lim_{n \to \infty} \left| \frac{2\cdot{6}\cdot{10}\cdots{4n+6}}{5\cdot{8}\cd ot{11}\cdots{3n+8}} \ \cdot \ \frac{5\cdot{8}\cdot{11}\cdots{3n+5}}{2\cdot{6}\cd ot{10}\cdots{4n+2}} \right|$ (simplified the brackets)

$\displaystyle {\color{white}\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_{n}}\right|} = \lim_{n \to \infty} \bigg| \frac{{\color{red} 2\cdot{6}\cdot{10}\cdots (4n+2)}(4n+6)}{{\color{blue}5\cdot{8}\cdot{11}\cdo ts (3n+5)}(3n+8)}$$\displaystyle {\color{white}.} \ \cdot \ \frac{{\color{blue}5\cdot{8}\cdot{11}\cdots{3n+5}} }{{\color{red}2\cdot{6}\cdot{10}\cdots{4n+2}}} \bigg|$ (expanded the inside of the dots)

Notice that the red's and blue's cancel out. Take the limit of what remains and the rest follows (see Mathstud's post).