Question

Determine whether the series is absolutely convergent, conditionally convergent, or divergent.

Answer 1

\[\frac{ 1 }{ 3 }+\frac{ 1*8 }{ 3*6 }+\frac{ 1*8*15 }{ 3*6*9 }+\frac{ 1*8*15*22 }{ 3*6*9*12 }+...\]

Answer 2

I assume the denominator will be 3n, but not sure about numerator

Answer 3

Close, the denominator of the \(n\)th term is a product, not just \(3n\). More like \(\displaystyle\prod_{k=1}^n3k\).

The numerator is adding multiples of \(7\) to the first term, which can also be written as a product, namely \(\displaystyle\prod_{k=1}^n(1+7(k-1))\).

So your series has the somewhat closed form
\[\sum_{n=1}^\infty\frac{\displaystyle\prod_{k=1}^n3k}{\displaystyle\prod_{k=1}^n(1+7(k-1))}\]

Answer 4

Sorry, that should be flipped... I meant
\[\sum_{n=1}^\infty\frac{\displaystyle\prod_{k=1}^n(1+7(k-1))}{\displaystyle\prod_{k=1}^n3k}\]

Answer 5

Using n, it would be \[\frac{ 7n }{ 3n }\] using the degree terms and neglecting the negligible terms when taking the limit?

Answer 6

so lim as n approaches inf = 7/3, which is greater than one, thus series is DIV?

Answer 7

To determine whether the series converges or not can be done using the ratio test.

The \((n+1)\)th term would be \(\dfrac{\displaystyle\prod_{k=1}^{n+1}(1+7(k-1))}{\displaystyle\prod_{k=1}^{n+1}3k}\), and the \(n\)th term would be \(\dfrac{\displaystyle\prod_{k=1}^n(1+7(k-1))}{\displaystyle\prod_{k=1}^n3k}\).

To use the ratio test, then, you need to show that the limit of the quotient of the \((n+1)\)th term and the \(n\)th term is less than \(1\). Fortunately, we're dealing with consecutive products, which means we can reduce the limit computation to something simpler:
\[\lim_{n\to\infty}\left|\frac{\dfrac{\displaystyle\prod_{k=1}^{n+1}(1+7(k-1))}{\displaystyle\prod_{k=1}^{n+1}3k}}{\dfrac{\displaystyle\prod_{k=1}^n(1+7(k-1))}{\displaystyle\prod_{k=1}^n3k}}\right|=\lim_{n\to\infty}\left|\frac{\dfrac{\displaystyle\prod_{k=n+1}^{n+1}(1+7(k-1))}{\displaystyle\prod_{k=n+1}^{n+1}3k}}{\dfrac{1}{1}}\right|=\lim_{n\to\infty}\frac{1+7((n+1)-1)}{3(n+1)}\]

Answer 8

\[=\lim_{n \rightarrow \infty} \frac{ 7n+8 }{ 3n+3 }=\lim_{n \rightarrow \infty} \frac{ 7n }{ 3n }=7/3>1\]
Thus DIV by Ratio test?

Answer 9

Yes, that's correct

Answer 10

Thanks much!

Answer 11

yw