![Find The Value Of Sum (n/2^n) [duplicate]](https://netmathematik.de/image/find-the-value-of-sum-n-2n-duplicate.jpeg "Find The Value Of Sum (n/2^n) [duplicate]")
The problem of finding the value of the infinite series ∑ (n/2^n) from n=1 to infinity is a fascinating example of how calculus can be used to solve seemingly complex problems. This series, often encountered in the study of infinite series and calculus, has a finite sum, which we can determine using a clever trick involving differentiation and geometric series. Let's explore the steps involved in finding the value of this sum.
Understanding the Series
The series ∑ (n/2^n) represents an infinite sum of terms where each term is obtained by dividing the index 'n' by 2 raised to the power of 'n'. The first few terms of this series are:
- n = 1: 1/2
- n = 2: 2/4 = 1/2
- n = 3: 3/8
- n = 4: 4/16 = 1/4
- ...
It's not immediately obvious what the sum of these infinite terms will be. To tackle this, we'll employ a combination of calculus and series manipulations.
The Approach: Differentiation and Geometric Series
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Introducing a Function: Let's define a function f(x) = ∑ (x^n / 2^n) from n=0 to infinity. This function is a geometric series with a common ratio of x/2.
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Geometric Series Formula: We know that the sum of an infinite geometric series is given by:
S = a / (1 - r)where 'a' is the first term and 'r' is the common ratio. In our case, a = 1 and r = x/2. Therefore:
f(x) = ∑ (x^n / 2^n) = 1 / (1 - x/2) = 2 / (2 - x) -
Differentiation: Now, let's differentiate both sides of the equation with respect to x:
f'(x) = d/dx [2 / (2 - x)] = 2 / (2 - x)^2 -
Finding the Sum: Our goal is to find the sum of the series ∑ (n/2^n). Notice that the derivative of f(x) gives us a series that looks remarkably similar:
f'(x) = ∑ (n * x^(n-1) / 2^n)If we set x = 1, we get:
f'(1) = ∑ (n / 2^n)This is precisely the series we want to find the sum of!
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Final Calculation: Substitute x = 1 into the expression for f'(x):
f'(1) = 2 / (2 - 1)^2 = 2
Therefore, the sum of the infinite series ∑ (n/2^n) from n=1 to infinity is 2.
Verification
While we arrived at this answer using a clever technique, it's always a good idea to verify our result. One way to do this is by looking at the partial sums of the series. As we add more terms, the sum should approach the value of 2.
For instance, the sum of the first 10 terms of the series is approximately 1.999, and the sum of the first 20 terms is approximately 1.99999. This pattern reinforces our conclusion that the series indeed converges to 2.
Conclusion
The infinite series ∑ (n/2^n) may seem daunting at first, but by cleverly applying differentiation and the properties of geometric series, we found its exact value to be 2. This problem highlights the elegance and power of calculus in tackling problems involving infinite sums and demonstrating how seemingly complex problems can be solved through a combination of techniques. Find the value of sum (n/2^n) becomes a straightforward calculation when we use these tools.
