Question

Need series help!

Answer 1

My Un = 5^n/n!

Answer 2

I have 2 more
2n^3+5/4n^5+1

And
(1/2^n )+1/3^(n+1)

Answer 3

I need to find whether they converge or diverge

Answer 4

the limit test is silent for for #1
so try ratio test as there is a possibility for easier simplification due to exponents and factorials

Answer 5

Yes I tried D-Alembert Ratio test and ended up getting n+1/r

Answer 6

whats r ?

Answer 7

sorry 5

Answer 8

I was trying with a generalized question..r^n/n!..same thing

Answer 9

ohkay :)
(n+1)/r is just a ratio right ? whats the limits as \(\large n \to \infty\) ?

Answer 10

and that ratio doesn't look correct, check once...

Answer 11

\[\LARGE \lim_{n \rightarrow \infty } \frac{U_{n+1}}{U_n} = \frac{n+1}{5}= \infty\]

Answer 12

if its really > 1, then the series diverges.

Answer 13

\[\LARGE \lim_{n \rightarrow \infty } \frac{U_{n+1}}{U_n} = \frac{5^n}{n!} \times \frac{(n+1)!}{5^{n+1}} =\]
and n+1 factorial = (n+1)n!

Answer 14

you have flipped the ratio

Answer 15

ohh sorry D: so its 5/n+1 then its 0

Answer 16

yes :) since its <1 for ANY r, the series converges for all r

Answer 17

ohh thanks next one!!

Answer 18

wait

Answer 19

what if its n!/r^n ? then its infinite?

Answer 20

if its infinity then its treated as divergent? no "finite answer" condition?

Answer 21

exactly! if the sum if infinity/swings up and down(cosx), we say the series is divergent

Answer 22

alright :O
\[\LARGE \frac{2n^3+5}{4n^5+1}\]

Answer 23

the series \(\large \sum\limits_{n=1}^{\infty}\sin(nx)\) diverges eventhough the sum can never be greater than 1

Answer 24

amazingg

Answer 25

so infinite sum is not the only way that a series can become divergent

Answer 26

i see!

Answer 27

for #2,

familair with p-series test ?

Answer 28

yep

Answer 29

Yep

Answer 30

\[0\lt \frac{2n^3+5}{4n^5+1}\lt \dfrac{2n^3+5}{4n^5} \lt \dfrac{1}{n^2}\]

Answer 31

so we will take \[\Large V_n = \frac{1}{n^2} \]
and take
\[\LARGE \lim_{n \rightarrow \infty} \frac{U_n}{V_n} =?\]

Answer 32

I thought so but ended up nowhere :/

Answer 33

since 1/n^2 series converges,
we an use comparison test

Answer 34

if you ignore "1" then we get the answer but why will we do so?!

Answer 35

what do you mean ignore 1 ?

Answer 36

\[0\lt \frac{2n^3+5}{4n^5+1}\lt \dfrac{2n^3+5}{4n^5} \lt \dfrac{1}{n^2}\]
you reduced 4n^5+1 to 4n^5

Answer 37

\[\LARGE \lim_{n \rightarrow \infty} \frac{U_n}{V_n} =\frac{2n^5 + 5n^2}{4n^5+1}\]

Answer 38

because \(\large \dfrac{1}{n+1}\lt \dfrac{1}{n}\)

Answer 39

this is the final thing if I do it my way

Answer 40

what exactly are you trying to do with the ratio ?
which test are you applying ?

Answer 41

ratio test....if Un/Vn is non 0 and finite number then Un behaves as Vn so if the ratio gave a finite answer I would've said that Un will be convergent since Vn is convergent(p series with n=2)

Answer 42

but I'm getting infinite here so my test fails..in any case If I take limit n tending to infinity I would get 0

Answer 43

ohh wait..limit is 4/5 in my method..it works ! :P

Answer 44

so Vn is convergent hence Un is convergent :O :O

Answer 45

step by step.

say \(\large a_n = \dfrac{2n^3+5}{4n^5+1} \)

the first test to try is limit test,
but the limit test is silent here because \(\lim \limits_{n\to \infty } a_n = 0 \)

Answer 46

uhh limit is 1/2...ive got the answer

Answer 47

a clarifying question :
is your second series exactly like below ?

\[\large \sum\limits_{n=0}^{\infty} \dfrac{2n^3+5}{4n^5+1}\]

Answer 48

yes

Answer 49

what are you trying with Un and Vn 's ?

Answer 50

what test is that ?

Answer 51

it seems we both are talking about two different tests

im not getting what exactly you're doing

Answer 52

its called Comparison test.
it states that:
If 2 positive term series \(\Large \sum_{}^{} U_n \) and \(\Large \sum_{}^{} V_n \) are such that \(\LARGE \lim_{n \rightarrow \infty} \frac{U_n}{V_n}\) is non 0 and finite then \(\Large \sum_{}^{} U_n \) behaves as \(\Large \sum_{}^{} V_n \) does,that is ,either both converge or diverge.

Answer 53

we assume Vn according to our own convenience,mostly its a P-series!!
here I took 1/n^2 i.e p series with p=2(p>1) therefore converges

Answer 54

there is still one q left :(

Answer 55

Ahh I see now ! looks perfect !!!

Answer 56

\[\large \sum\limits_{n=0}^{\infty}\dfrac{(1/2^n )+1}{3^{n+1}}\]

Answer 57

like this ?

Answer 58

I'm given a series.......
\[\Large \frac{1}{2}+\frac{1}{3^2}+\frac{1}{2^3}+\frac{1}{3^4}......\infty\]
so my Un should be
\[\Large \sum \frac{1}{2^n} + \frac{1}{3^{n+1}}\] i guess

Answer 59

sum of two converging series is convergent

Answer 60

your original series is a sum of two converging geometric series, right ?

Answer 61

I thought of applying D-Alembert test so I found Un+1
\[\Large U_{n+1}= \sum \frac{1}{2.2^n} + \frac{1}{3.3^{n+1}}\]
\[\Large \lim_{n \rightarrow \infty} \frac{U_{n+1}}{U_n}=\frac{3.{\cancel{3^{n+1}}+2.\cancel{2^n}}}{2.2^{n} 3.3^{n+1}} \times \frac{\cancel{3^{n+1}}+\cancel{2^n}}{\cancel{2^n} \cancel{ 3^{n+1}}}\]

Answer 62

i didn't know that :o can't we prove it like this?

Answer 63

\[\large \left( \frac{1}{2}+\frac{1}{2^3}+\frac{1}{2^5}+\cdots\right)+\left(\frac{1}{3^2}+\frac{1}{3^4}+\cdots\right)\]

Answer 64

i messed up the calculation lol

Answer 65

sure we can prove it using ratio test also

Answer 66

but ur general term is not correct (it doesnt matter though)

Answer 67

i knew that was coming :P

Answer 68

try below :

\[\large \sum \limits_{n=1}^{\infty} \dfrac{1}{2^{2n-1}} + \dfrac{1}{3^{2n}}\]

Answer 69

\[\sum U_{n+1}=\large \sum \limits_{n=1}^{\infty} \dfrac{1}{2^{2n}} + \dfrac{1}{3^23^{2n}}\]

Answer 70

take the ratio and somehow show that it will be less than 1

Answer 71

you're taking ratio of n+1th TERM by the nth TERM

NOT the ratio of SUMS.

Answer 72

\(\large U_n = \dfrac{1}{2^{2n-1}} + \dfrac{1}{3^{2n}}\)


\(\large U_{n+1} = \dfrac{1}{2*2^{2n}} + \dfrac{1}{9*3^{2n}}\)

Answer 73

\[\large \lim \limits_{n\to \infty} \dfrac{U_{n+1}}{U_n} = \lim \limits_{n\to \infty} \dfrac{\dfrac{1}{2*2^{2n}} + \dfrac{1}{9*3^{2n}}}{\dfrac{1}{2^{2n-1}} + \dfrac{1}{3^{2n}}}\]

Answer 74

Clearly this doesn't look that pleasant

Answer 75

i was doing the same :/

Answer 76

we could have simply concluded the series converges using the fact that :

sum of two converging series converges

Answer 77

okay nvm lets forget this one,ill ask my teacher if i can do that way :P

Answer 78

\[\Large \sum 2^{-n-(-1)^n}\]
no idea about this one..i sense Cauchy though

Answer 79

can i take power 1/n directly like that?

Answer 80

you want to take nth root ?

Answer 81

yes

Answer 82

yes root test looks good to me

Answer 83

\[\Large (U_n)^{\frac{1}{n}}= 2^{-n-(-1)}\] is it correct?

Answer 84

i guess not :P

Answer 85

\[\Large (U_n)^{\frac{1}{n}}= 2^{(-n-(-1)^n)/n}\]

Answer 86

\[\Large (U_n)^{\frac{1}{n}}= 2^{(-1-(-1)^n/n)}\]

Answer 87

\[\Large \lim_{n \rightarrow \infty}{(-n-(-1)^n)^{\frac{1}n}}\]
so do we need to solve this?

Answer 88

\[\Large (U_n)^{\frac{1}{n}}= \frac{1}{2}2^{-(-1)^n/n}\]

Answer 89

now take the limit

Answer 90

what's going on ? :|

Answer 91

\[\large \lim\limits_{n\to \infty}(U_n)^{\frac{1}{n}}= \lim\limits_{n\to \infty}\frac{1}{2}2^{-(-1)^n/n} = \frac{1}{2}\]

Answer 92

since the limit is < 1, the series converges by root test