Question

If r < 1 and positive and m is a positive integer , show that .

$(2m+1)r^m(1-r) \ \textless \ 1-r^{(2m+1)}$.

And also show that nrⁿ is indefinitely small when n is indefinitely great.

Answer 1

Ans:1 Geometric Progression :

Steps and Understanding :
1) We will write the sum of GP with
first term = 1
common ratio = r ( 0< r < 1 )
no. of terms (2m +1).

Like :
$1 + r + {r}^{2} + .... + {r}^{2m} = \frac{1 - {r}^{2m + 1} }{1 - r}$


2) Then, since all terms are positive and distinct we will use concept of
AM > GM
As shown in pic.

3) After calculation we will get required answer.
------------------------

Ans: 2 This is trivial question but Enthusiastic.

Observing the question, we just need to prove,
lim n tends to infinity : nr^n = 0
As here, when n becomes indefinitely great then, then it's limit approaches to 0 or practically value which is very close to it.
That, value is indefinitely small than value of n which is indefinitely great.


There are two methods one of which is logical, and another is Mathematical.

Method : 1
1) Let the function be
$f(n) = n {r}^{n} \: \: \: \: where \: 0< r < 1$

2) Then,
Replace r with 1/t where t> 1 as 0<r < 1.
Then,
We have to find
lim n tends to infinity :
$\frac{n}{ {t}^{n} }$

3) This form of limit is inf/inf .
So, we will apply L-Hospital rule to check out its limiting value.
After solving. it's value approaches to 0.



Method - 2 :
1) We know that Exponential functions like x^n where 0< x <1 decreases more faster than linear functions like n.
So, nx^n tends to indefinitely small value when n becomes indefinitely great.

Answer 2

Step-by-step explanation:

(2m+1)r

m

(1−r)<1−r

2m+1

⇒(2m+1)r

m

<

1−r

1−r

2m+1

The term in RHS is the sum of a GP 1,r,r

2

,..,r

2m+1

and r

m

being the middle term.

If No of terms × Middle term < Sum of terms, then ∣r∣<1.

Here, r is positive ⇒0<r<1

So, r

n

would be very small quantity for high values of n.

Hence, nr

n

is also indefinitely small for indefinite large value of n.