Question

\[\sum_{n=1}^{\infty} (-1)^{n} \frac{2^{n}n!}{5*8*11*...*(3n+2)}\]
\[\lim_{n \rightarrow \infty} \frac{5*8*11*...*(3n+2)}{2^{n}n!} \frac{2^{n+1}(n+1)!}{5*8*11*...*(3(n+1)+2)}\]
\[\lim_{n \rightarrow \infty} \frac{(3n+2)}{2^{n}} \frac{2^{n+1}(n+1)}{(3(n+1)+2)}\]
\[\lim_{n \rightarrow \infty} \frac{2^{n+1}}{2^{n}} \frac{(n+1)(3n+2)}{(3(n+1)+2)}\]
\[\lim_{n \rightarrow \infty} 2\frac{(n+1)(3n+2)}{(3(n+1)+2)}\]

Answer 1

hope im on the right track

Answer 2

Do you want to compute the sum or prove convergence, divergence?

Answer 3

convergence or divergence

Answer 4

\[\lim_{n \rightarrow \infty} 2\frac{3n^{2}+5n+2}{3n+5}\]

Answer 5

Limits is infinity

Answer 6

because the numerator is a degree higher than the denominator?

Answer 7

I think you can cancel the (3n+2) in the numerator cause in the denominator the sequence should be 5*8*11*...*(3n+2)*(3n+5), don't you?

Answer 8

I considered that ...but is it legal?

Answer 9

I think it's legal, that makes the limit 2/3 < 1 and the serie converges

Answer 10

\[\lim_{n \rightarrow \infty} \frac{(3n+2)}{2^{n}} \frac{2^{n+1}(n+1)}{(3(n+1)+2)}\]
show me the math

Answer 11

please

Answer 12

because according to eliassaab it's divergent

Answer 13

Before that on the second step

Answer 14

or am is misunderstanding @eliassaab

Answer 15

Ok, maybe I'm wrong then

Answer 16

Your series is convergent. Try to see why.

Answer 17

why?

Answer 18

lolz sorry guys...i'm not quite understanding

Answer 19

\[\lim_{n \rightarrow \infty} 2\frac{3n^{2}+5n+2}{3n+5}\]
am i right up to this point?

Answer 20

I think you're not, I mean on second step, we have in the numerator 5*8*11*...*(3n+2) right?

Answer 21

yes

Answer 22

Ok, in the denominator we have the x sub(n+1) and it is 5*8*11*...*(3n+2)*(3n+5)

Answer 23

\[
\lim_{n->\infty } \left |\frac {a_{n+1}}{a_n} \right|=\frac 23
\]

Answer 24

We can cancel the (3n+2) terms, that what I'm trying to say, but as I said above, maybe I'm wrong

Answer 25

So it is convergent.

Answer 26

\[\lim_{n \rightarrow \infty} 2\frac{(n+1)(3n+2)}{3(n+1)+2}\]

@eliassaab does it matter that the numerator is a degree higher than the denominator?

Answer 27

ok you are both right according to my solution manual...

Answer 28

but i'm still having trouble with the elimination part I guess

Answer 29

sigh...the mystery

\[\lim_{n->\infty } \left |\frac {a_{n+1}}{a_n} \right | =
\lim_{n->\infty } \frac {a_{n+1}}{a_n} =\\
\lim_{n->\infty } \frac {2^{n+1} (n+1)!}
{5. 8 \cdots (3n+1)(3n+4)} \frac{5. 8 \cdots (3n+1)} {2^{n} n!}\\\]
what happened here
\[ =\frac {2 n}{3n+4}\to \frac 2 3\]

Answer 30

Did you get it?

Answer 31

i get this part
\[=\frac {2 n}{3n+4}\to \frac 2 3\]

Answer 32

By the ration test, the series converges.

Answer 33

@knock @eliassaab plz don't hate me lolz ...i'm not intentionally trying to be so stubborn...i guess i'll just stare at it a little longer

Answer 34

is it 2/3 or -2/3

Answer 35

You have to take the absolute value of the ratio. So it is 2/3

Answer 36

i understand it's convergent but i guess i'm having trouble with the algebra part eliminating form the numerator and denominator...(which is the easiest part of this problem)

Answer 37

What is (n+1)!/n!

Answer 38

n+1

Answer 39

Imagine that you have this sequence 5*8*11...(3n+2), 3n+2 verify the terms because for n=1 it's 5, for n=2 it's 8 and so
well then you see the diference between each number it's 3, so the next term after 3n+2 should be 3n+5,
as all terms are multipliying to each other we can cancel all of the and we have 3n+5 in the denominator

Answer 40

I had a small misprint.

Answer 41

ok now we can cross out 3n+1

Answer 42

VICTORY!!!!

Answer 43

We can cross out 5.8, ... (3n+1)

Answer 44

yep yep now I can see it

Answer 45

Great ^^

Answer 46

Great

Answer 47

thanks! saved my life you two

Answer 48

Yep but I think the terms are 3n+2 aren't they? Cause 3n+4 doesn't makes 5*8*11... for n=1,2,...

Answer 49

It is 3n +5

Answer 50

Ok, agree then yw guys

Answer 51

\[
\lim_{n->\infty } \left |\frac {a_{n+1}}{a_n} \right | =
\lim_{n->\infty } \frac {a_{n+1}}{a_n} =\\
\lim_{n->\infty } \frac {2^{n+1} (n+1)!}
{5. 8 \cdots (3n+2)(3n+5)} \frac{5. 8 \cdots (3n+2)} {2^{n} n!}\\
=\frac {2( n+1)}{3n+5}\to \frac 2 3
\]

Answer 52

makes sense now

Answer 53

The numerical sums of your series is -0.269723