Question

show that k=1 sigma k tends to infinity ((1/k)-(1/k+2)) =3/2

Answer 1

So,


\[\sum_{k=1}^{\infty} \frac{1}{k}-\frac{1}{k+2}= \left(\frac{1}{1}-\frac{1}{3}\ \right)+\left(\frac{1}{2}-\frac{1}{4}\ \right)+\left(\frac{1}{3}-\frac{1}{5}\ \right)+\left(\frac{1}{4}-\frac{1}{6}\ \right).... \]

So as you can see after some time, the third set of terms in parenthesis starts to cancel out the negative terms. As this goes on these negative terms all fade out with all the positive terms except the first two, thus, \[1+\frac{1}{2}=\frac{3}{2}\]

Answer 2

heloo any one see my qus.. plz help me.

Answer 3

bro can u explain now what u did?

Answer 4

Ask me questions.

Answer 5

whole term whrer going.?

Answer 6

Term by term is going to 0. As you can see the fractions are getting smaller.

Answer 7

thisi is telescoping series..?

Answer 8

Yes.

Answer 9

how can calculte the sum of this series..?>??

Answer 10

Simple, but subtle.

First, look at it as sets of numbers.
So, look at them in groups. The first parenthesis is the first group, then the second parenthesis is the second group, and third pare...you get the point. Scroll up and look at them then write me back when you have.

Answer 11

any formula calculate the telescoping series sum?

Answer 12

Yes, add all the terms up until one less than the telescoping term.

Answer 13

There's a formula, but it looks ugly.

Answer 14

in my qustion how to use formula ?

Answer 15

What is the telescoping term?

Answer 16

i dont know..:(

Answer 17

Back to square one.

"
Simple, but subtle.

First, look at it as sets of numbers.
So, look at them in groups. The first parenthesis is the first group, then the second parenthesis is the second group, and third pare...you get the point. Scroll up and look at them then write me back when you have.
"

Answer 18

formula.?

Answer 19

Here you go:

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

Where \[H_k\] is the Kth harmonic number.

Answer 20

what is 1/n waht is n.?

Answer 21

\[H_k=\sum_{t=1}^k \frac{1}{t}\]

Answer 22

For the first term, or group of parenthesis, its n=1 for the the second one is 2 for the third is 3 for the foutrh is.....you get the point, its all the integers and as you go on looking through the set of parenthesis you find bigger values for n.

Answer 23

Hk/k Hk=1/t and what is K ?

Answer 24

It defines the telescoping terms within the series....unless you want to start from square one, none of these variables and constants will make sense, even if you have the formula.

Answer 25

so bro in my qus what is Hk and what is K?

Answer 26

K=2 and Hk is 1+1/2=3/2

Answer 27

how..?

Answer 28

Magic....Seriously, if you don't start form square one it wont make sense anyway.

Answer 29

plz bro explain now..:(

Answer 30

I got to go, sorry.