Question

The following series converge by the ratio test. Use summation by parts, \(\sum_{k=1}^{n} a_{k}\left(b_{k+1}-b_{k}\right)=\left[a_{n+1} b_{n+1}-a_{1} b_{1}\right]-\sum_{k=1}^{n} b_{k+1}\left(a_{k+1}-a_{k}\right)\), to find the sum of the given series. $$ \left.\sum_{k=1}^{\infty} \frac{k}{2^{k}} \text { (Hint: Take } a_{k}=k \text { and } b_{k}=2^{1-k} .\right) $$

Short answer

The sum of the series \( \sum_{k=1}^{\infty} \frac{k}{2^{k}} \) is \( 2 \).

Step-by-step solution

Identify the Terms

We start by identifying the terms in the given series. We have the series \( \sum_{k=1}^{\infty} \frac{k}{2^{k}} \). We are given the hint to take \( a_k = k \) and \( b_k = 2^{1-k} \). This sets up the terms for summation by parts.

Apply the Summation by Parts Formula

The formula given is:\[\sum_{k=1}^{n} a_{k}(b_{k+1}-b_{k}) = [a_{n+1} b_{n+1}-a_{1} b_{1}] - \sum_{k=1}^{n} b_{k+1}(a_{k+1}-a_{k})\]We need to find \( b_{k+1} - b_k \) and \( a_{k+1} - a_k \). Let's calculate these next.

Calculate Differences

First, find \( b_{k+1} - b_{k} \):\[ b_{k} = 2^{1-k} \quad \text{so} \quad b_{k+1} = 2^{-k} \]Thus,\[ b_{k+1} - b_{k} = 2^{-k} - 2^{1-k} = -2^{1-k} \]Now, find \( a_{k+1} - a_{k} \):\[ a_{k} = k \quad \text{so} \quad a_{k+1} = k+1 \]\[ a_{k+1} - a_{k} = 1 \]

Substitute and Simplify

Substitute these differences back into the summation by parts formula:\[\sum_{k=1}^{n} k(2^{-k} - 2^{1-k}) = [a_{n+1} b_{n+1} - a_1 b_1] - \sum_{k=1}^{n} 2^{-k} \cdot 1\]Calculate endpoints:\[ [a_{n+1} b_{n+1} - a_1 b_1] = [(n+1)2^{-n} - 1]\]Thus:\[\sum_{k=1}^{n} \frac{k}{2^{k}} = [(n+1)2^{-n}] - [1] - \sum_{k=1}^{n} 2^{-k}\]

Simplify the Result

The series \( \sum_{k=1}^{n} 2^{-k} \) is a geometric series with the sum:\[ S = \frac{1 - (\frac{1}{2})^n}{1 - \frac{1}{2}} = 2(1 - 2^{-n}) = 2 - 2^{1-n} \]Plug this back into the formula:\[ (n+1)2^{-n} - 1 - (2 - 2^{1-n}) = (n+1)2^{-n} - 1 - 2 + 2^{1-n} \]\[ = (n+1)2^{-n} - 2 + 2^{1-n} + 2^{1-n} \]Observe the behavior as \( n \to \infty \).The terms involving \(2^{-n}\) and \(2^{1-n}\) approach zero, thus simplifying to \( 2 \).

Conclusion

As the limits for \( n \to \infty \) are evaluated, the final sum of \( \sum_{k=1}^{\infty} \frac{k}{2^{k}} = 2 \).

Key concepts

Ratio Test

Understanding whether a given infinite series converges or diverges is key in calculus. One powerful tool for this purpose is the Ratio Test. It is primarily used for series where each term is a ratio of the preceding term. To apply the ratio test, consider an infinite series \( \sum a_k \). The test involves calculating the limit

• \( L = \lim_{{k \to \infty}} \left| \frac{a_{k+1}}{a_k} \right| \)

Depending on the value of \( L \), the series may:

• Converge absolutely if \( L < 1 \)
• Diverge if \( L > 1 \)
• Remain inconclusive if \( L = 1 \)

For the original problem, the ratio test is already satisfied, affirming the convergence of the series. This allows us to then find the series' sum using other techniques like summation by parts.

Geometric Series

A geometric series is a type of series with a constant ratio between successive terms. It can be expressed as \( a + ar + ar^2 + ar^3 + \ldots \), where \( a \) is the first term and \( r \) is the common ratio. One of the powerful aspects of geometric series is that they have a neat formula for their sum:

• For an infinite geometric series where \( |r| < 1 \), the sum \( S = \frac{a}{1 - r} \).

In the context of the given problem, we used this concept to find the sum of the series \( \sum_{k=1}^{n} 2^{-k} \). This geometric series has a first term \( a = \frac{1}{2} \) and a ratio \( r = \frac{1}{2} \), leading to an expression for its partial sum. This step is part of simplifying the overall series.

Infinite Series

An infinite series involves the sum of terms that continue indefinitely. Each series has its own "convergence behavior," determining if the sum approaches a finite limit.

• An infinite series is written as \( \sum_{k=1}^{\infty} a_k \).
• The sum is said to be convergent if it adds up to a finite number.
• If it doesn't, the series is divergent.

Evaluating such series to find their sum often involves using various techniques like the ratio test, known series formulas, or summation by parts.
In our problem, by applying these methods, particularly summation by parts supported by the behavior of series sums, we calculated that \( \sum_{k=1}^{\infty} \frac{k}{2^k} = 2 \). Here, as \( n \to \infty \), we ensured the decreasing nature of terms sharply converges to the finite sum.