Question

Can someone help me in solving this question ?

Answer 1

\[\sum_{i = 1}^{\infty} (\cfrac{1}{T_i})\]

Answer 2

Where, \[T_{r+1} = T_r + 30 + 18(r-1) = T_r + 18r + 12\]

Answer 3

Also ,
\[T_1 = 10\]

Answer 4

I know this sum will converge. But, I can't find out how to start solving this.

Answer 5

@satellite73

Answer 6

I tried to use invariance principle.

Answer 7

But couldn't find anything useful.

Answer 8

@ganeshie8

Answer 9

@mathmale

Answer 10

Some questions to help you get started: Is this a geometric series, or not? How would you know?

What are the first several terms of this series?

What are the first several sums of this series? Do the sums seem to approach some limit?

Answer 11

(1) No its not a geometric series. The ratio between two consecutive terms is not the same.

(2) 1/10, 1/40, 1/88, 1/154, ...

(3) Sum does approach to a limit as the numbers in later will start approaching to zero.

Answer 12

I can think of something.

Answer 13

I tried it in this way:

\[\cfrac{1}{10} + \cfrac{1}{40} + \cfrac{1}{88} + ...\]
\[= \cfrac{1}{3} (\cfrac{1}{2} - \cfrac{1}{5}) + \cfrac{1}{3} (\cfrac{1}{5} - \cfrac{1}{8}) +...\]
\[= \cfrac{1}{3}(\cfrac{1}{2} - \cfrac{1}{(2+3r)})\]

where \[r \rightarrow \infty \]

Answer 14

So,

sum approaches \(\cfrac{1}{6}\). Is this right ?

Answer 15

I agree that this does not seem to be a geometric series.

Your approach is a very intelligent one and I applaud you on being ab le to recognize the presence of a telescoping series.

Answer 16

I find out using this:

\[T_{r+1} - T_r - 18r = 12\]
\[(3r+2)(3r+5) - (3r-1)(3r+2) - 18r = 12\] (I tried finding out \(T_{r+1}\) and \(T_r\) using first few terms)

The second equation is true.

So, there I got it.

Thanks @mathmale , for your help and your words of appreciation. :)

Answer 17

After all... he is gone on me...! :P

Answer 18

Haha haha !! ;)

Answer 19

Really, LITTLE bro !! O.O I didn't know that and didn't even realise that.

Answer 20

Congrats, Vish. Great work. Seems as though you have a lot of promise in Maths!

Answer 21

No, seriously, I taught you about this all.... You forgot everything :/ I'm depressed..! ;)

Answer 22

Thanks everyone for your help .

Thanks @mathmale