Question

test the convergence
Xn= 1+1/2!+1/3!.....1/n!

Answer 1

see its bounded '\(1 \le X_n \le 1+c \) :P
check c as uper bound
also its monotonic \(a_n > a_{n-1}\)
so
bounded +monotonic =convergence :D

Answer 2

awhhh
okay okay

Answer 3

u dnt need to find c though :D

Answer 4

since 1/n! goes to zero
then its bounded on half closed set

Answer 5

1/n also goes to zero..does that tell you anything about 1+1/2+1/3+...+1/n as n goes to infinity

Answer 6

yep , its also converge

Answer 7

no...it is the harmonic series...it diverges

Answer 8

bounded +monotonic :O

Answer 9

if the series is bounded and monotonic then it converges

Answer 10

you are just bounding a term and not the whole series

Answer 11

ok let me use the def :)

Answer 12

there's a criterion which involves \(\lim_{n\to\infty}\frac{a_{n+1}}{a_n}\)..

Answer 13

A sequence X=\(\large (x_n)\) in R is said to be converge to \(x\in R\), or x said to be a limit of \(x_n\) if for every \(\epsilon >0 \) there exist a natural number\( K( \epsilon )\) such that for all n\(\ge \) \( K( \epsilon )\) the terms \(x_n \) satisfy \( |x_n-x | \) < \(\epsilon \)
if a sequence has a limit , we say that the sequence is convergent ,if it has no limit , we say that the sequence is divergent

Answer 14

wew !
according to this i say its conv.
also 1/n conv
what do u say ?

Answer 15

yes @reemii that is one way to do it if one does not recognize the sum

Answer 16

the original does converge...I know the sum.

\[\sum_{n=1}^{\infty}\frac{1}{n}\]diverges

so they way you are thinking is wrong

Answer 17

why lol

Answer 18

http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)

Answer 19

>.>
hehe its series and i prove for sequence ,
sorry

Answer 20

o_o i hv 4 more assignments to write

Answer 21

in 2 hours

Answer 22

so what do i write here >_<

Answer 23

harmonic sequence diverges
so it does not converge?

Answer 24

1/n div
1/n! conv
but lets reason it again , dnt write the comment above

Answer 25

okay?

Answer 26

.-.

Answer 27

\( \large x_n =\sum \frac {1}{n!} \)
its convergent ,
enough to show by Induction n^2 <n! for n\(\ge\) 4
but first u need to know 1/n^2 is conv. also ( u can prove it :) )
so
\(\large 0<\frac {1}{n!}<0<\frac {1}{n^2}\) for n>=4
>.>

Answer 28

\(\large x:=\frac {1}{n!} \)
\(\large y:=\frac {1}{n^2} \)
then when \(n\ge 4\) we have


\(\large 0\le \frac {x_n}{y_n }=\frac {n^2 }{n!} <\frac {1}{n-2} \rightarrow 0 \)

Answer 29

.-.
i cant remember this fully lol but i guess its correct hehe

Answer 30

then by \(\Huge \color{red}{\text { Limit Comparison Test }}\) Apllies the conv
"u should know this test :D "

Answer 31

done :D

Answer 32

bravo ^^

Answer 33

ikr ^^
thx to @Zarkon to notice that i was working for sequence :D