Question

Try it guyz

Answer 1

Solve for x :-

\[\huge x = 1+\frac{ 1 }{ 3+\frac{ 1 }{ 2 + \frac{ 1 }{ 3 + \frac{ 1 }{ 2 + ..... \infty } } } }\]

Answer 2

\[\large x = 1+\frac{1}{3 + \frac{1}{1+x} }\]

Answer 3

that nested thingy from the indian guy

Answer 4

31-7 is the answer.

Answer 5

intellingent you have ripped out the question @ganeshie8

Answer 6

31/7

Answer 7

oh sorry
i thought something else.

Answer 8

answer is

\[\huge \sqrt{\frac{ 5 }{ 3 }}\]

Answer 9

nice work. @ganeshie8

Answer 10

his name is ramanujan @ikram002p

Answer 11

Whose name?

Answer 12

Is that a joke?

Answer 13

Yeah
Srinivasa Ramanujan
the name speaks for itself

Answer 14

but this is not that case i think

Answer 15

is the value of x a approximation @ganeshie8 @ikram002p

Answer 16

How can we give a finite value to the infinite sequence i don't understand

Answer 17

how about geometric series ?

Answer 18

yeah i don't get

Answer 19

\[1+\frac{1}{2} + \frac{1}{4} + \frac{1}{8}+\cdots = 2\]

Answer 20

so it is a approximation right

Answer 21

it is exact

Answer 22

it tends to 2

Answer 23

yes, it approaches 2

Answer 24

this is a good question , its approaches but exact xD

Answer 25

LOL

Answer 26

im serous :D

Answer 27

2 can approach 2 ?

Answer 28

@ikram002p Are you coming back to my Binomial theorem question?

Answer 29

yeah

Answer 30

\[\sum \limits_{k=0}^{\infty} \frac{1}{k^2} = 2\]

the series exactly equals 2

Answer 31

the sequence of partial sums approach 2

Answer 32

:)

Answer 33

\[\sum \limits_{k=0}^{\text{large n}} \frac{1}{k^2} \approx 2\]

Answer 34

\[\sum \limits_{k=0}^{\infty} \frac{1}{k^2} =2\]

Answer 35

here why we use approach or tend when its infinite :-
\(\sum \limits_{k=0}^{\infty} \frac{1}{k^2} = \lim_{n \to \infty } \sum \limits_{k=0}^{n} \frac{1}{k^2} \)

Answer 36

\[\sum \limits_{k=0}^{\infty} \frac{1}{k^2} = \lim_{n \to \infty } \sum \limits_{k=0}^{n} \frac{1}{k^2} = 2\]

Answer 37

it makes no sense to say the infinite sum approaches 2 right ?

Answer 38

hehe its a concept xD its not something i made
as n tend to infinite the sum tend to 2
ehh idk why u dont got it , when u take open sets they teach u this

Answer 39

in real analysis

Answer 40

the example i always like
sum 1/2^n

Answer 41

are you suggesting the series approximately equals 2 ?

Answer 42

nope

Answer 43

im saying approaches xD or tend ( little difference )
u can say about it exact if ur not doing a prove of some analysis ( js)

Answer 44

ok

Answer 45

k

Answer 46

my head rejects the phrase "the infinite series approaches 2"

it just makes no sense to me

Answer 47

you can say "the partial sums approach 2 as you increase n"
but the infinite series in question is always identically equal to 2

Answer 48

if you get what im saying... idk if usage of words like this is right... but thats how i make sense of it

Answer 49

u have a point its just a concept thingy u know , there is no such infinite value that makes the error=0 , that is why proper meaning or understanding would make sense

Answer 50

the limit of a sequence, if it exists is a fixed value. its not running as you increase/decrease n... the limit is not a moving target to say it is approaching something hmm