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Magazine Chapter 5: Problem 172
Use the estimate \(R_{N} \leq \int_{N}^{\infty} f(t) d t\) to find a bound for the remainder \(R_{N}=\sum_{n=1}^{\infty} a_{n}-\sum_{n=1}^{N} a_{n}\) where \(a_{n}=f(n) .\) $$ \sum_{n=1}^{100} n / 2^{n} $$
Short Answer
Expert verified
The upper bound for the remainder is very small due to the exponential decay, specifically \( \frac{101}{\ln(2)^2 \cdot 2^{100}} \).
Step by step solution
01
Identify the Function
In this problem, the given sequence is represented as \( a_n = \frac{n}{2^n} \). We can express this as a function of a continuous variable \( t \) for the integral, so \( f(t) = \frac{t}{2^t} \).
02
Define the Remainder Estimate
We're tasked with finding an upper bound for the remainder \( R_N = \sum_{n=1}^{\infty} a_n - \sum_{n=1}^{N} a_n \). The formula \( R_N \leq \int_{N}^{\infty} f(t) \, dt \) will be used to calculate this bound.
03
Set Upper Limit for N
In the problem, we're asked to evaluate \( \sum_{n=1}^{100} \frac{n}{2^n} \). Therefore, \( N = 100 \) and the remainder \( R_N \) is the part of the series we haven't summed up to infinity, beyond 100.
04
Evaluate the Integral
Compute the integral \( \int_{100}^{\infty} \frac{t}{2^t} \, dt \). This integral looks complicated, but can be approached using integration by parts or recognized as a decay factor exponential type integration for large \( t \).
05
Apply Integration Techniques
For integration by parts, let \( u = t \) and \( dv = \frac{1}{2^t} dt \), then \( du = dt \), and integrate \( dv \) to get \( v = \int \frac{1}{2^t} dt = -\frac{1}{\ln(2) \cdot 2^t} \).
06
Compute using Integration by Parts
Using integration by parts: \( \int_{N}^{\infty} \frac{t}{2^t} \, dt = \left. \left(-\frac{t}{\ln(2) \cdot 2^t}\right) \right|_{N}^{\infty} + \int_{N}^{\infty} \frac{1}{\ln(2) \cdot 2^t} dt \), the first part evaluates to zero as both terms go to zero at \( \infty \).
07
Simplify Remaining Integral
The second term simplifies to \( \frac{1}{\ln(2)} \int \frac{1}{2^t} dt = -\frac{1}{\ln(2)^2 \cdot 2^t} \). Evaluate this from 100 to infinity to find its contribution to the bound.
08
Calculate Bound
The bound evaluates to approximately \( \frac{101}{\ln(2) \cdot 2^{100} \cdot \ln(2)}\), which simplifies further to a small numeric value due to the \( 2^{100} \) term, forming a small upper bound.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Infinite Series
An infinite series is a sum of infinite terms that follow a specific sequence. It can be thought of as a way to add up numbers that continue indefinitely. For example, in our exercise, we're considering the series \( \sum_{n=1}^{\infty} \frac{n}{2^{n}} \).
Unlike finite sums, infinite series do not reach a fixed total immediately, and instead often approach a limit. Infinite series can
Unlike finite sums, infinite series do not reach a fixed total immediately, and instead often approach a limit. Infinite series can
- converge to a specific value, meaning the sum approaches a certain number as more terms are added;
- or diverge, meaning the sum continues to grow indefinitely.
Integration by Parts
Integration by parts is a technique derived from the product rule of differentiation and is used to integrate products of functions. It is best remembered with the formula \[\int u \, dv = uv - \int v \, du, \]where you choose which part of the integral should be \( u \) and which should be \( dv \).
This method is particularly helpful when dealing with functions like \( \frac{t}{2^t} \), as seen in our problem. You choose
This method is particularly helpful when dealing with functions like \( \frac{t}{2^t} \), as seen in our problem. You choose
- \( u = t \), making \( du = dt \),
- and \( dv = \frac{1}{2^t} \, dt \), leading to \( v = -\frac{1}{\ln(2) \cdot 2^t} \).
Exponential Decay
Exponential decay describes processes that decrease at a rate proportional to their current value. It is characterized by functions like \( \frac{1}{2^t} \).
This type of function is important in our remainder estimation problem, particularly because it approaches zero very quickly as \( t \) increases.
Exponential decay is important because
This type of function is important in our remainder estimation problem, particularly because it approaches zero very quickly as \( t \) increases.
Exponential decay is important because
- it simplifies many calculations due to its predictable decline rate,
- and it allows us to make approximations, especially when considering large \( t \) values.
Upper Bound Estimation
Upper bound estimation in the context of infinite series involves approximating the maximum potential sum that remains when considering the sum beyond a certain point.
In our exercise, the upper bound for the remainder \( R_N \) is estimated using \( \int_{N}^{\infty} f(t) dt \), which provides a way to estimate how much of the series is left unsummed.
An upper bound
In our exercise, the upper bound for the remainder \( R_N \) is estimated using \( \int_{N}^{\infty} f(t) dt \), which provides a way to estimate how much of the series is left unsummed.
An upper bound
- tells us the worst-case scenario, where the sum could go if it continued past a certain point,
- and helps confirm that the remainder is small, making numerical calculations easier to manage.
