Question

\[\sum_{n=1}^{\infty} (-1)^{n+1} \frac{n^{2} 2^{n}}{n!}\]

ratio test?

Answer 1

\[\frac{(n+1)^{2}2^{(n+1)}}{(n+1)!} * \frac{n!}{n^{2}2^{n}}\]

Answer 2

\[(-1)^{n+2}{(n+1)^22^{n+1}\over(n+1)!}\cdot\frac1{(-1)^{n+1}}{n!\over n^22^n}\]

Answer 3

it becomes
\[\frac{n^2}{2n+2}\]

Answer 4

TuringTest...where did you get the \[(-1)^{n+2}\]
from?

Answer 5

sorry flit up

Answer 6

\[\frac{2n+2}{n^2}\]

Answer 7

\[(-1)^{n+2}{(n+1)^22^{n+1}\over(n+1)!}\cdot\frac1{(-1)^{n+1}}{n!\over n^22^n}\]\[(-1){2n^2+4n+2\over n^2(n+1)}\]@MathSofiya the term after (-1)^(n+1) is (-1)^(n+2)

Answer 8

oops!.... thanks

Answer 9

\[(-1)^{n+2}{(n+1)^22^{n+1}\over(n+1)!}\cdot\frac1{(-1)^{n+1}}{n!\over n^22^n}\]\[(-1){2n^2+4n+2\over n^2(n+1)}=-2{\left({n+1}\over n^2\right)}\]does anybody disagree with that?
cinar where did the - go ?

Answer 10

whatever, it's absolute value anyway

Answer 11

and it tends to zero

Answer 12

yeah that's good..

Answer 13

I think for alternating series ,,,
proving the elements tends to zero as n tends to infinity
and each term is decreasing term {of absolute sequence } would be enough.

however absolute convergence implies convergence.

Answer 14

check this out
http://tutorial.math.lamar.edu/Classes/CalcII/AlternatingSeries.aspx

Answer 15

why does my solution manual ignores the \[(-1)^{n+1}\] ?

Answer 16

Turning Test please fan me you are AWESOME!!!!

Answer 17

Hmm ... it's an alternating series,
proving the convergence of absolute term ... also implies convergence

Answer 18

exactly...that's what Paul online notes does too

Answer 19

If \[ \frac{n^{2} 2^{n}}{n!}\] is convergent then
\[ (-1)^n \frac{n^{2} 2^{n}}{n!} \]
is obviously convergent.

Though it can trouble you sometime . for eg
\( \frac 1 n \) is divergent while \( (-1)^n \frac 1n \) is convergent.

Answer 20

\[ b_{n}\] is \[\frac{n^{2} 2^{n}}{n!}\]
so we can ignore
\[(-1)^{n+1}\] when calculating convergence using the ratio test
...correct?

Answer 21

Yes you can, but do not always use this method. Why? see my above post.
it's better if you show
\[ \frac{(n+1)^22^{n+1}}{(n+1)!} \leq \frac{(n)^22^{n}}{(n)! }\]
and
\[ \lim_{x \rightarrow \infty } \frac{(n)^22^{n}}{(n)! } = 0\]

Answer 22

you cannot use that method on
\[ \sum_{1}^\infty \frac{(-1)^n} n \]

Answer 23

Ok
\[\frac{(n+)^{2}2^{(n+1)}}{n^{2}2^{n}} * \frac{n!}{(n+1)!}\]

I'm not the best at doing limits...so lets see
the fraction on the right becomes \[\infty\]?

Answer 24

plz don't give up on me just yet

Answer 25

was in not you whom I talked to yesterday about dividing factorials?

Answer 26

yes
and
the term become n+1

Answer 27

or n in this case

Answer 28

right, keep going

Answer 29

ok n+1

Answer 30

1/(n+1) in this case

Answer 31

n-> \[\infty\] so 1/1

Answer 32

no limits yet, we're not done simplifying

Answer 33

\[ \frac{(n+1)^{2}2^{(n+1)}}{n^{2}2^{n}} * \frac{n!}{(n+1)!} = \left ( 1+ \frac 1n \right )^{2}2^{1} \frac{1}{n+1} \]
taking limit you get zero which is less than 1 so the series converges absolutely.

Answer 34

ok so I have \[ \frac{(n+)^{2}2^{(n+1)}}{n^{2}2^{n}}\] * \[\frac{1}{n+1}\]

Answer 35

right, so how can you simplify this part\[\frac{(n+1)^{2}2^{(n+1)}}{n^{2}2^{n}}\]?

Answer 36

Interesting value
http://www.wolframalpha.com/input/?i=sum [n^2+2^n%2Fn!%2C+1%2C+infinity]

Answer 37

^weird...

Answer 38

this one is interesting too
http://www.wolframalpha.com/input/?i=sum [%28-1%29^%28n%2B1%29+n^2+2^n%2Fn!%2C+1%2C+infinity]

Answer 39

\[\frac{(n+1)^{2}2^{(n+1)}}{n^{2}2^{n}}\]
=

ok I can't do the elimination we did with factorials...so I don't really know....
Please give me hint

Answer 40

never forget that\[\large\frac{x^a}{x^b}=x^{a-b}\]

Answer 41

the fastest way to type latex is copying. Right click on latex ... show Math as Tex commands

Answer 42

right click->show math as->tex commands
then copy and paste, and enclose in brackets \[\.] (without the dot . )

Answer 43

\[\frac{(n+1)^{2}2^{(n+1)}}{n^{2}2^{n}}\]
=

\[\frac{(n+1)^{2} 2}{n^{2}}\]

Answer 44

yes :) times...?

Answer 45

\[\frac{(n+1)^{2} 2}{n^{2}}\] * \[\frac{1}{n+1}\]

Answer 46

i seeeeeeeee

Answer 47

\[\frac{(n+1)2}{n^{2}}\]

Answer 48

and the limit of that as n tends to infinity is...?

Answer 49

1/ infty =0

Answer 50

WOW!!! 43 MIN LATERRR!!!

Answer 51

guess better later than never ey

Answer 52

notice how the division makes the n drop out mostly in cases like this, so when you get the hang of it you'll see certain parts coming.
for instance, that 2^(x+1)/2^x=2 is really normal in these problems, so next time I bet you'll do it a lot faster
like how you learned factorials :)

Answer 53

Thanks

Answer 54

welcome!