So this one is confusing me:
limit of n as it approaches infinity of (n{SIGMA}i=1) (1/(2^i)
*the n is on top of the sigma notation and i=1 is at the bottom.
I understand that 'i' would equal (n(n+1))/2 however the algebra from there confuses me.
The sum is
1/(2^i)
= 1/2^1 + 1/2^2 + 1/2^3 + ...
= 1/2 + 1/4 + 1/8 + ...
You will recognize that as a geometric series, with
S = (1/2)/(1 - 1/2) = 1
so a is the first term in the series and r is the term that is being reused over and over again ?
in the equation a/(1-r)
naturally, since that is the formula for the sum of a G.P.
To solve this problem, let's break it down step by step:
Step 1: Rewrite the given expression.
The expression is the sum of 1 divided by 2 raised to the power of i, where i ranges from 1 to n. We can rewrite this as:
(nā[i=1] (1/(2^i)))
Step 2: Simplify the given sum.
In order to simplify the sum, we can use the formula for the sum of a geometric series:
ā(k=1 to n) ar^(k-1) = a((1-r^n)/(1-r))
Here, a = 1 and r = 1/2.
So, the sum becomes:
(n * (1 - (1/2)^n) / (1 - 1/2))
Simplifying further, we get:
2n * (1 - (1/2)^n)
Step 3: Take the limit as n approaches infinity.
To find the limit of this expression as n approaches infinity, we need to examine the behavior of each term.
- The term "2n" grows linearly with n.
- The term "(1/2)^n" decreases exponentially as n gets larger.
As n approaches infinity, the exponential term becomes negligible compared to the linear term.
Therefore, taking the limit as n approaches infinity, we can ignore the exponential term, and the expression simplifies to:
Limit of n as it approaches infinity of (2n) = infinity
So, the limit of the given expression as n approaches infinity is infinity.