Question

show\quad that\quad the\quad series\quad \frac { 1 }{ 2 } +\frac { 1 }{ 4 } -\frac { 1 }{ 8 } +\frac { 1 }{ 16 } +\frac { 1 }{ 32 } -\frac { 1 }{ 64 } +...is\quad convergent\quad and\quad that\quad its\quad sum\quad lies\quad in\quad [-1,1].

Answer 1

\[show\quad that\quad the\quad series\quad \frac { 1 }{ 2 } +\frac { 1 }{ 4 } -\frac { 1 }{ 8 } +\frac { 1 }{ 16 } +\frac { 1 }{ 32 } -\frac { 1 }{ 64 } +...is\quad convergent\quad and\quad that\quad its\quad \sum\quad lies\quad \in\quad [-1,1].\]

Answer 2

this is mathematical analysis problem.and that its sum lies in [-1,1].

Answer 3

show that the serie is greater than zero and less than one.

Answer 4

the series is absolutely convergent, it's natural that it converges.

Answer 5

can you simplify the series.i meant to say that can u write it as a sum.

Answer 6

take the absolute values of the series, you notice that it is a geometric series that converges to 1. hence your series must be less than 1. you also know that your series is greater than 0. hence it must converge between 0 and 1

Answer 7

show me the steps pls.

Answer 8

\[ \sum_{n=1}^{\infty } a_n \le \sum_{n=1}^\infty |a_n| = \sum_{n=1}^\infty \frac {1}{2^n} = 1 \]

Answer 9

notice that 1/4 - 1/8 = +1/8
you conclude that the series is (monotonically) increasing and is bounded above by 1. this gives your converging sequence in (0,1)

Answer 10

THANK YOU @experimentX .

Answer 11

you are welcome.

Answer 12

PROVE\quad THAT\quad 2{ n }^{ 2 }\ge { (n+1) }^{ 2 }\quad for\quad n\ge 3\quad mathematical\quad analysis\quad problem.

Answer 13

\[PROVE\quad THAT\quad 2{ n }^{ 2 }\ge { (n+1) }^{ 2 }\quad for\quad n\ge 3\quad mathematical\quad analysis\quad problem.\]

Answer 14

use induction

Answer 15

and also post your question in different post.

Answer 16

Expanding the left side gives
2(k2+2k+1)≥?[(k+1)+1]2
right?

Answer 17

Expanding the left side gives
2(k2+2k+1)≥?[(k+1)+1]2
right?

Answer 18

but what about rhs (right hand side.) ? i don't know how to prove from here.

Answer 19

hold on a sec .... buddy, type inside latex. also use different post if you want it to get more attention.

Answer 20

Sure :)
Let's consider whether or not
\[2(k+1)^{2}≥? [(k+1)+1]^{2}\]

Answer 21

yes

Answer 22

Expanding the left side gives
\[2(k ^{2}+2k+1)≥?[(k+1)+1]^{2}\]
right?

Answer 23

but what about rhs (right hand side.) ?

Answer 24

don't expand the left side .. only expand the right side ... also with respect to (k+1)^2 + 2(k+1) + 1 so you may use induction

Answer 25

how to deal with greater than or equation can you explain a bit more with the steps pls .

Answer 26

i really wish you had posted in another post ... I was freaking busy right now
\[ 2(k+1)^{2}≥ [(k+1)+1]^{2} \\
2(k+1)^2 \ge (k+1)^2 + 2 (k+1) + 1 \\
(k+1)^2 \ge 2(k+1) + 1 \]
this is always true for \[ (k+1) >= 2 \]

Answer 27

or just
\[ (k+1)(k - 1) \ge 1 \\
k^2 \ge 2 \]
earlier you assumed k>2 so this is true.

Answer 28

\[(k+1^{})^{2}≥2(k+1)+1 \]

Answer 29

this is always true for (k+1)>=2 how ?