Question

k=1sigma k tends to infinity (1/(k+1)(k+3))
how can i find this series is converges

Answer 1

If the nth term tends to zero as n tends to infinity it converges

Answer 2

plz solve the steps.?

Answer 3

Wts the nth term?

Answer 4

i dont know. how to find nth term of this sequence

Answer 5

just replace k by n

Answer 6

and then put n tending to infinity

Answer 7

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

Answer 8

if the nth term tends to 0 the divergence test is inconclusive

Answer 9

put the limit then bottom value is infiny and 1/infinity is zero..? so can i find converges

Answer 10

are u sure,...im rusty on this topic

Answer 11

integral test, if you get a number out of the unbound integral it is divergenct and if not then its not

Answer 12

see i dunno that much

Answer 13

zzrocker plz can help me in this question..?

Answer 14

can you do the integral of f(x) from 0 to infinity?

Answer 15

err 1 to infinity?

Answer 16

k=1 to infinity

Answer 17

hellow ru there/?

Answer 18

take the integral from 1 to infinity yes

Answer 19

can u solve it..?

Answer 20

in my question what is the closed form of the nth partil sum.?

Answer 21

http://www.wolframalpha.com/input/?i=integral+%281%2F%28k%2B1%29%28k%2B3%29%29+from+1+to+infinity

Answer 22

Im not sure what that is asking you, i think sigma 1->n f(x)

Answer 23

k=1 sigma k tends to infity

Answer 24

my question is how can i find closed form of the nth partial sum and whehter the series in coverges.?

Answer 25

hellow any one see my queston

Answer 26

We can't give you the answer sherazzzzz, we can only help you solve it.

Answer 27

ok dear little bit help me plz

Answer 28

if you do the integral you will see the series does not converge, because the integral does not converge. and sigma 1->n f(x) is how you write it

Answer 29

sigma 1->n f(n)

Answer 30

my book ans says converges to 1/2

Answer 31

but how..? i dont know..:(

Answer 32

yes I did the wrong integral
http://www.wolframalpha.com/input/?i=integral+%281%2F%28%28k%2B1%29%28k%2B3%29%29%29+from+1+to+infinity

Answer 33

you know that it converges because the inttegral converges. do you understand that?

Answer 34

yes but how to find closed form of the nth partial sum.. this is direcect converges not to find nth partila sum

Answer 35

hi buddy u understand my problem??

Answer 36

no, I cant read what you wrote.

Answer 37

find the closes form of the nth partial sum and determin wheter the seriers in converges?

Answer 38

Find the closed form of the nth partial sum and determine whether the series converges?

Answer 39

Do you understand that I have answered you about "if the series converges" ?

Answer 40

how to find a closed form for the nth partial sum?

Answer 41

yes borther but problem is how to find closed form for the nth partila sum?

Answer 42

I think that is just saying use sigma notation? to write the function series expansion in terms of n with sigma notation>

Answer 43

yes bro

Answer 44

ok so I have said this many times as well

sigma from 1 to n of f(n)

Answer 45

\[\sum_{k=1}^{\infty} 1/(k+1)(k+3)\]

Answer 46

1/((K+1)(K+3)) but yes, if that is what they are asking you to do, then that is the answer:)

Answer 47

yar converges to 1/2 how..? con u know it.?

Answer 48

is (k+1)(k+3) both in the denominator?

Answer 49

yes

Answer 50

um there are many ways to see what it converges to, what are you doing in class?

Answer 51

are you doing geometric series?

Answer 52

brother i study in virtual university..

Answer 53

no ;(

Answer 54

Hey guys, what about comparision test?

\[\sum_{k}^{\infty} 1/(k+1)(k+3) = \sum_{k}^{\infty} 1/(k^2+4k+3)\]

\[1/(k^2+4k+3) < 1/k^2 \] so comparing with an armonic series as 1/k^2, as the bigger converge the smaller converge too

Answer 55

he knows it converges, but how to get the sum is the question?

Answer 56

brother not to find comparison ttest just find closed from of nth partial sum

Answer 57

are you hulk hogan?

Answer 58

bro plz help me ..:( plz how to find closed form of the nth partial sum

Answer 59

I dont know. if its not geometric i dont know how to do it.

Answer 60

but your book says 1/2 and wolfram says 5/12

Answer 61

yes bro its not a geometrc serires..:(

Answer 62

telescopic series

Answer 63

you can make 1/(k+1)(k+3)=(1/2)[(1/(k+1)) - (1/(k+3))]

Answer 64

and now you can solve as a telescopic serie

Answer 65

O yeah , I sort of remember those. rewrite it as two series and look for canceling pattern.

Answer 66

yeah I've done the serie and the sum is 5/12 so this is the way

Answer 67

may I write it?

Answer 68

yeah same here, i was gonna scan it but please go ahead:)

Answer 69

first expand then write as 1/2sum(1/(x+1)) - 1/2sum(1/(x+3)) =
1/2(1/2+1/3+1/4+1/5......) - 1/2(1/4+1/5.....) cancel like terms and get 1/2(1/2+1/3) = 5/12

Answer 70

the book is wrong:)

Answer 71

ok so as I said above you can make 1/(k+1)(k+3)=(1/2)[(1/(k+1)) - (1/(k+3))];

now if you make k=1 the firs term is

1/2-1/4

if you give k=2

1/3-1/5;
1/4-1/5 (k=3);
...

1/k - 1/(k+2) ; (k-1 term)
1/(k+1) - 1/(k+3) ; (k term)

so if you cancel terms you have that S=1/2+1/3- 1/(k+2)- 1/(k+3);
you compute the limit when k--> infinity and S=5/6 but cause we factor out 1/2 we have
(1/2)*S=(1/2)*(5/6)=5/12

Answer 72

there ya go two dif roads lead to same answer:)

Answer 73

it's not a quite clear explanation but I don't know how to write in LaTex yet ^^

Answer 74

agree, I think the book is wrong

Answer 75

does this all make sense sharazzzz?

Answer 76

bah I think hes gone lol

Answer 77

haha maybe, I hope he'll see this, anyway it's solved

Answer 78

thanks all of you.. but my question not solve. because i ask the question how to find closed form for a nth partial sum of the series..?

Answer 79

dear 1/2 k say aaay..?

Answer 80

? I have told you the answer 13 times