Problem 35 - Calculus: One and Several Variables

Question

(a) Assume that \(d_{k} \rightarrow 0\) and show that \(\sum_{k=1}^{x}\left(d_{k}-d_{k+1}\right)=e d_{1}\) (b) Sum the following series: (i) \(\sum_{i=1}^{\infty} \frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k(k+1)}}\) (ii) \(\sum_{k=1}^{\infty} \frac{2^{L}+1}{2 k^{2}(k+1)^{2}}\)

Step-by-Step Solution

Answer

In summary: (a) We showed that \(\sum_{k=1}^{x}\left(d_{k}-d_{k+1}\right)=ed_1\), which is true as the sequence \(d_k\) converges to 0. (b) For the given series, we found the following sums: (i) \(\sum_{i=1}^{\infty} \frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k(k+1)}} = 1\) (ii) \(\sum_{k=1}^{\infty} \frac{2^{L}+1}{2 k^{2}(k+1)^{2}} = \frac{\pi^4}{96} + 2^{L-1}\left(\frac{\pi^4}{96}\right)\)

Step 1: Rewrite the summation using properties of sums

We can simplify the summation by breaking it into two separate sums. Since both of the sums have the same indices, we can rewrite the term inside the sum as the difference between the two sums: \[\sum_{k=1}^{x}\left(d_{k}-d_{k+1}\right) = \sum_{k=1}^{x} d_k - \sum_{k=1}^{x} d_{k+1} \]

Step 2: Adjust the index of the second sum

To make the indices match for both sums, we can do a change of variable in the second sum. Let \(j = k+1\), so \(k = j-1\). Now the summation becomes: \[\sum_{k=1}^{x} d_k - \sum_{j=2}^{x+1} d_j \]

Step 3: Subtract the sums

Now we can subtract the terms of the two summations, remembering that as k goes to infinity, \(d_k\) goes to 0: \[(d_1 - d_2) + (d_2 - d_3) + \cdots + (d_x - d_{x+1}) = d_1 - d_{x+1}\] Since \(d_{k}\rightarrow 0\) as \(k\rightarrow \infty\), we get: \(d_1 - d_{x+1} = ed_1\) #Part (b): Summing the given series# #(i) Summing the first series#

Step 1: Simplify the series

We can start by simplifying the terms in the series. We see that we can cancel out the square root terms in the numerator and the denominator: \[\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k(k+1)}} = \frac{1}{\sqrt{k}+\sqrt{k+1}}\]

Step 2: Telescoping sum method

Now, we can rewrite the series using telescoping sum method. We can notice that the series given to us converges since the terms go to zero as k goes to infinity. \[\sum_{k=1}^{\infty} \frac{1}{\sqrt{k}+\sqrt{k+1}} = 1\] #(ii) Summing the second series#

Step 1: Simplify the series

We can notice a pattern in the given series which will help us simplify the terms: \[\frac{2^{L}+1}{2 k^{2}(k+1)^{2}} = \frac{1}{k^2(k+1)^2} + \frac{2^L}{2k^2(k+1)^2}\]

Step 2: Rewrite the series using partial fraction decomposition

Now, we can use partial fraction decomposition to rewrite the series as a difference of fractions: \[\frac{1}{k^2(k+1)^2} = \frac{A}{k^2} + \frac{B}{(k+1)^2}\] Solving for A and B, we get \(A = (k+1)^2\) and \(B = -k^2\), therefore, \[\frac{1}{k^2(k+1)^2} = \frac{(k+1)^2 - k^2}{k^2(k+1)^2}\]

Step 3: Telescoping sum method

We can use telescoping sum method to find the sum of the series. We need to expand the second term in the simplified expression before summing: \[\sum_{k=1}^{\infty} \frac{2^L}{2k^2(k+1)^2} = 2^{L-1}\sum_{k=1}^{\infty} \frac{1}{k^2(k+1)^2}\] Now, we can sum the series (since it converges): \[\sum_{k=1}^{\infty} \frac{1}{k^2(k+1)^2} = \frac{\pi^4}{96}\] and \[\sum_{k=1}^{\infty} \frac{2^L+1}{2 k^{2}(k+1)^{2}} = \frac{\pi^4}{96} + 2^{L-1}\left(\frac{\pi^4}{96}\right)\]

Key Concepts

Telescoping Series

A telescoping series is a special type of infinite series. This happens when consecutive terms cancel each other out. As a result, the series simplifies quickly. For example, in series like \( (a_1 - a_2) + (a_2 - a_3) + \ldots + (a_n - a_{n+1}) \), most terms cancel each other. The series boils down to just a few terms like \( a_1 - a_{n+1} \). This is beneficial in finding limits and sums of infinite series. In telescoping, you can often see only the first and the last terms matter. This makes it elegant for finding sums efficiently without infinite sums being needed. For instance, if \( a_n \) approaches zero, then the sum simplifies further.

Partial Fraction Decomposition

Partial fraction decomposition is a method for breaking down fractions into simpler parts. It is useful in calculus for integrating complex rational functions and summing series. For example, consider the fraction \( \frac{1}{k^2(k+1)^2} \). With partial fraction decomposition, we can express it as:

• \( \frac{A}{k^2} \)
• \( \frac{B}{(k+1)^2} \)

By solving for \( A \) and \( B \), we can simplify the problem. This makes it easier to integrate or to use in telescoping series. Learning to decompose a fraction helps to handle complex mathematical expressions. It's like breaking a puzzle into smaller parts to solve it more easily.

Infinite Series

An infinite series is the sum of infinitely many terms, written as \( \sum_{k=1}^{\infty} a_k \). These series can take various forms: geometric, arithmetic, or more complex ones. Convergence is key here. A series converges when its terms approach zero, resulting in a finite sum. For example, the series \( \sum_{k=1}^{\infty} \frac{1}{k^2} \) converges to a known value. Convergence is essential in mathematics as it ensures we can work with infinite sums meaningfully, and a convergent series has a "limit" or finite result even if it technically continues forever.

Mathematical Induction

Mathematical induction is a proof technique used to prove formulas true for all natural numbers. It involves two main steps. First, we prove the base case: show the statement is true for \( n=1 \). Second, we assume it holds for some \( n = k \) (the induction hypothesis). Then show the statement holds for \( n=k+1 \). This step-by-step method helps verify statements about sequences, series, and more. It's like climbing a ladder: first prove the bottom rung is safe, then if one step holds, another will. Mathematical induction offers a solid foundation for proving truths across all numbers without checking each case individually.