Question

the series form n=1 to infinity of :
2*5*8.......(3n-1)/2^n*n!

Answer 1

\[\sum_{1}^{\infty}2*5*8....(3n-1)/(2^n*n!)\]

Answer 2

I would suggest maybe trying to find a better way to write the numerator since as it stands it's sort of hard to see what to do with it.

Answer 3

Wait, are we trying to find the sum or trying to find if it converges or what?

Answer 4

sorry forgot to say that
Determine if the series converges conditionally , converges absolutely or diverges

Answer 5

Well, in your own words explain what the difference is between conditional and absolute convergence.

Answer 6

\[\large=\sum_{n=1}^{\infty}\left[ \prod_{k=1}^{n}{3k-1\over2k} \right]\]

Answer 7

well conditional is if u take the absolute value of the series and its divergent , but without the absoulute it is convergent, then that is conditional convergent
amd absolute convergent is
when u take the absolute and both series, the one with the absolute and the one without it both converge

Answer 8

\[\large=\sum_{n=1}^{\infty}\left[ \prod_{k=1}^{n}{3-\frac1k\over2} \right]\]

Answer 9

\[\large=\sum_{n=1}^{\infty}{2 \over 2}\cdot{2\frac12 \over 2}\cdot{2\frac23 \over 2}\cdot{2\frac34 \over 2}...\cdot{2+\frac{ n-1 }{ n } \over 2}\]

the seuence diverges ... so the series diverges

Answer 10

\[\sum_{n=1}^{\infty}\frac{2\times5\times8\times\cdots\times(3n-1)}{2^nn!}\]
Using the ratio test, you have
\[\lim_{n\to\infty}\left|\frac{2\times5\times8\times\cdots\times(3n-1)(3(n+1)-1)}{2^{n+1}(n+1)!}\times\frac{2^nn!}{2\times5\times8\times\cdots\times(3n-1)}\right|\\
\lim_{n\to\infty}\left|\frac{3(n+1)-1}{2(n+1)}\right|\\
\lim_{n\to\infty}\left|\frac{3n+2}{2n+2}\right|\\
\frac{3}{2}\]
which means the series diverges because the limit is greater than 1.

Answer 11

ok thanks, can u explain how they canceled

Answer 12

\[\lim_{n\to\infty}\left|\frac{2\times5\times8\times\cdots\times(3n-1)(3(n+1)-1)}{2^{n+1}(n+1)!}\times\frac{2^nn!}{2\times5\times8\times\cdots\times(3n-1)}\right|\\

\lim_{n\to\infty}\left|\frac{\cancel{2\times5\times8\times\cdots\times(3n-1)}(3(n+1)-1)}{2\cancel{2^{n}}(n+1)\cancel{n!}}\times\frac{\cancel{2^n}\cancel{n!}}{\cancel{2\times5\times8\times\cdots\times(3n-1)}}\right|\]