Question

Can help me on this?

Answer 1

Question in attachment.

Answer 2

e

Answer 3

any way the terms cancel in pairs so youre left with the first term and the secon the last two terms you add in any time. this gives you 1/0 ie infinity.

Answer 4

Can you show the working?

Answer 5

just add em up. the second term of the first set and the first of the second set will cancel. this goes on. you're left with only the first term.

Answer 6

like 1/n - 1/(n|1) + 1/(n+1) - 1/(n+1|1) ... so on. here | stands for a + .. which then makes the middle terms cancel always.

Answer 7

How does the first part of the question connected to the 2nd one?

Answer 8

Is there a formula or something for this? I didnt learn this before.

Answer 9

it helps you expand each term of the series. | as such means nothing. the first part is there to tell you that | ~ +. not a formula. it's about match the symbols.

Answer 10

anyway, the first part is also a key to translate 1/n.(n|1) into terms we can understand. ok ?

Answer 11

I still cant get it.sorry.

Answer 12

i meant "| means +"

Answer 13

fine. i'll write it for you..

\[\sum_{n-1}^{\infty} 2\div (n \times(n|1)) = 2\times \sum_{n-1}^{\infty} 1\div (n \times(n|1))\]

Answer 14

now the first part of the question tells you what, 1÷(n×(n|1)) is. put that in. and expand. youll get terms like the pair i wrote above..1/n - 1/(n|1) + 1/(n+1) - 1/(n+1|1).. the middle ones cancel. get it?

Answer 15

since n>= 1 the first term is 1/0 which is infinity.

Answer 16

how do you get the pair?

Answer 17

c'mon , write it out for n & then for n+1 . you have four terms now. the middle two cancel. so youre left with the first term (which never cancels) & the last term of the pair you add last.

Answer 18

How you expand btw?

Answer 19

using the first part.. "now the first part of the question tells you what, 1÷(n×(n|1)) is. put that in. and expand. youll get terms like the pair i wrote above..1/n - 1/(n|1) + 1/(n+1) - 1/(n+1|1).. the middle ones cancel. get it?
"

Answer 20

can u show how to expand?

Answer 21

i will .

1/n.(n|1) = 1/n - 1/n|1 . here this makes a lot of sense if you put n|1 as n+1 . now that reads
1/n.(n+1) = 1/n - 1/n+1

Answer 22

i think you know how to treat summation signs ?

Answer 23

hmm nope. how to treat the summation sign?