Question

series review - convergence

Answer 1

\[\large \sum \limits_{n=2}^{\infty } \dfrac{n^2+3n+2}{7^n}\]

Answer 2

the sequence converges as \(\large \lim \limits_{n\to \infty} a_n = 0\)
how to go about testing the series ?

Answer 3

u wanna find the sum or only determine converge /diverge

Answer 4

just determine whether its converges/diverges

Answer 5

wolfram is using ratio test, should i try the same ?

Answer 6

\[\large \lim \limits_{n\to \infty} \Bigg |\dfrac{a_{n+1}}{a_n}\Bigg| = \lim \limits_{n\to \infty} \Bigg |\dfrac{((n+1)^2+3(n+1)+2)/7^{n+1}}{(n^2+3n+2)/7^n}\Bigg| \]

Answer 7

\[\large = \lim \limits_{n\to \infty} \Bigg |\dfrac{((n+1)^2+3(n+1)+2)}{7(n^2+3n+2)}\Bigg| \]

Answer 8

somehow this needs to be less than 1

Answer 9

for this i often use limit comparison test , ur series can be defided into 3 series right ?

Answer 10

\[\large = \Bigg |\dfrac{1}{7}\Bigg | \]
\[\large \lt 1\]

this turned out simple xD

Answer 11

yeah lets try limit comparison now :)

Answer 12

we need to choose a reference series to compare against ?

Answer 13

yep mmm

Answer 14

converge one

Answer 15

how about \(\large \dfrac{6^n}{7^n}\) ?

Answer 16

mmm we need to prove that
n^2+3n+2 <6^n

Answer 17

\[\large \sum \limits_{n=2}^{\infty } \dfrac{6^n}{7^n}\]

and

\[\large \sum \limits_{n=2}^{\infty } \dfrac{n^2 + 3n+2}{7^n}\]

Answer 18

n^2+3n+2 <= 6^n

will do right ?

Answer 19

yeah it work ^^

Answer 20

see how the comparison Test work
let X:=\((x_n)\) and Y:=\((y_n)\) be real sequences and suppose that for some K\(\in N\) we have
\(0\le x_n\le y_n \) for \(n\ge k\)
a) then the convergence of \(\sum y_n \) implies the convergence of \(\sum x_n\)
b) the divergence of \(\sum x_n\) implies the divergence of \(\sum y_n\)

Answer 21

interesting, so for some \(n \ge k\),
if \(0 \le x_n \le y_n\) , then \(x_n\) converges if \(y_n\) converges is it ?

Answer 22

so just by looking at this series , and define any other
convergence serieis such as 6^n/7^n > x_n
then its converge

Answer 23

yes ^_^

Answer 24

then we could have used 2^n/7^n also as reference series !!

Answer 25

mmm
if
n^2+3n+2> 2^2 then yes it works

Answer 26

2^n **

Answer 27

1.000000000001^n/7^n also would have worked !! sweet :3

Answer 28

haha :P

Answer 29

cuz, (1.0000000001)^n > an^2+bn+c for some n > k

Answer 30

you have helped me understand the comparison test clearly, thank you !

Answer 31

np :P

Answer 32

comparison test is the fastest test to see convergence :D
so do u think
\(\sum_{n=1}^{\infty}\frac{1}{n!}\) converge or diverge ?

Answer 33

lets see, i feel ratio test is ideal when dealing with factorials

Answer 34

ratio test for kids :P

Answer 35

ugh :/ ratio test still excites me

Answer 36

ok lol
what do u think is it conv/div and why :P

Answer 37

ratio test :

\[\large \lim \limits_{n\to \infty} \Bigg |\dfrac{a_{n+1}}{a_n}\Bigg| = \lim \limits_{n\to \infty} \Bigg |\dfrac{1/(n+1)!}{1/n!}\Bigg|\]

Answer 38

\[\large = \lim \limits_{n\to \infty} \Bigg |\dfrac{1}{n+1}\Bigg| \]

\[\large = 0\]

\[\large \lt 1\]

converges

Answer 39

?

Answer 40

yeah it work :)

Answer 41

what about comparison test

Answer 42

for comparison test its enough to know
n^2< n! , n>=4

hence
1/n^2 >1/n! >0 for n >=4

Answer 43

so its conv.

Answer 44

whoa yes !!

Answer 45

can we prove 1/n^2 converges quick ?

Answer 46

prove

\[\large \sum \limits_{n=1}^{\infty } \dfrac{1}{n^2}\]

converges

Answer 47

i know the proof using integral test, so lets try something elsse..

Answer 48

yes its p-series all
\(\Huge \sum_{1}^{\infty} \frac{1}{n^p}\) , s.t |p|>1
are converges

Answer 49

yeah can we go through the proof once

Answer 50

the sequence of partial sums is monotonic as all the terms are positive

Answer 51

how to show this is bounded

Answer 52

i like to write the theorm xD

let \(x_n\) be a sequence of nonnegative real numbers , Then the series \(\sum x_n\) converges if and only if the sequense S=\((S_k)\) of partial sums is bounded . In this case :-
\(\sum_{n=1}^{\infty } x_n=\lim(S_k)=sup[S_k :k \in N]\)

Answer 53

okay so we need to show that the sequence of partial sums is bounded

Answer 54

so as u said since partial sums are monotone
we need to show that some subsequensce of \(S_k\) is bounded :P

Answer 55

i used\(s_k\) def.. tell if u wanna me to write it down.

if \( k_1 := 2^1-1=1\) then \(S_{k_1}\) =1
if \(k_2:=2^2-1=3\) then \(S_{k_2}=3\)
then
\(s_{k}=\frac{1}{1}+(\frac{1}{2^2}+\frac{1}{3^2})<1+\frac{2}{2^2}=1+\frac{2}{2 }\)
and
if\( k_3:=2^3-1=7 \) then we have
\(S_{k_3}=S_{k_2}+(\frac{1}{4^2}+\frac{1}{5^2} +\frac{1}{6^2}+\frac{1}{7^2})<S_{k_2}+ \frac{4}{4^2}<1+\frac{1}{2}+\frac{1}{2^2}\)
with me so far ?

Answer 56

so if u keep going like this ull gonna note that
\(\large 0<S_{K_i<1}+\frac{1}{2}+(\frac{1}{2 })^2+. . . +(\frac{1}{2})^{j-1}\)

Answer 57

are you getting this ?

Answer 58

so looks like you're over estimating the 1/n^2 series right ?

Answer 59

and the over estimated series is just a geometric series with r = 1/2, so the upper bound is 2... okay got it :)

Answer 60

^_^ yes

Answer 61

now u show me how to prove it with integral test

Answer 62

integral test is tricial

Answer 63

*trivial

Answer 64

\[\large \sum \limits_{n=1}^{\infty} \dfrac{1}{x^p} \text{ converges } \iff \int \limits_1^{\infty } \dfrac{1}{x^p} dx \text{ converges } \]

Answer 65

\[\large \sum \limits_{n=1}^{\infty} \dfrac{1}{x^p} \text{ diverges } \iff \int \limits_1^{\infty } \dfrac{1}{x^p} dx \text{ diverges} \]

Answer 66

just evaluate the integral and analyze

Answer 67

\( \large \int \limits_1^{\infty } \dfrac{1}{x^p} dx = \lim_{x\rightarrow\infty } \dfrac{1}{x^p} -1=-1\)

Answer 68

nice hehehe i more like Real analysis so i try to imply that as much as i can

Answer 69

\[\large \int \limits_{1}^{\infty } \dfrac{1}{x^p}dx = \lim \limits_{t \to \infty } \int \limits_{1}^{t } \dfrac{1}{x^p}dx \]

Answer 70

\(p \ne 0\) :

\[\large = \lim \limits_{t \to \infty } ~\dfrac{x^{-p+1}}{-p+1}\Bigg |_1^t \]

Answer 71

\(p \ne 1\) *

Answer 72

yeah lol i was rushing :P

Answer 73

thx !

Answer 74

okay next module is on alternating series, wil post as a separate q

Answer 75

thank you for all the help :)

Answer 76

eshhh -_-