Write a MATLAB function file that computes the sum and makes a semilog-x plot of the value of this sum as a function of k for k = 1,2,…, 1000.

k S = \sum_{n=1} \frac{1}{n(n+1)} Write a MATLAB function file that computes the sum and makes a semilog-x plot of the value of this sum as a function of k for k = 1,2,…, 1000. Your numerical solution should suggest the value of this sum for k = \infty. Derive the value of the infinite sum.

The Correct Answer and Explanation is:

To solve this problem, we need to evaluate the sum SkS_kSk​ given by:Sk=∑n=1k1n(n+1)S_k = \sum_{n=1}^{k} \frac{1}{n(n+1)}Sk​=n=1∑k​n(n+1)1​

Step 1: Simplifying the Expression

The given term 1n(n+1)\frac{1}{n(n+1)}n(n+1)1​ can be decomposed using partial fractions:1n(n+1)=1n−1n+1\frac{1}{n(n+1)} = \frac{1}{n} – \frac{1}{n+1}n(n+1)1​=n1​−n+11​

Thus, the sum SkS_kSk​ becomes:Sk=∑n=1k(1n−1n+1)S_k = \sum_{n=1}^{k} \left( \frac{1}{n} – \frac{1}{n+1} \right)Sk​=n=1∑k​(n1​−n+11​)

This is a telescoping series, meaning that most terms cancel out. When expanded, we get:Sk=(1−12)+(12−13)+(13−14)+⋯+(1k−1k+1)S_k = \left( 1 – \frac{1}{2} \right) + \left( \frac{1}{2} – \frac{1}{3} \right) + \left( \frac{1}{3} – \frac{1}{4} \right) + \dots + \left( \frac{1}{k} – \frac{1}{k+1} \right)Sk​=(1−21​)+(21​−31​)+(31​−41​)+⋯+(k1​−k+11​)

After cancellation, the result simplifies to:Sk=1−1k+1S_k = 1 – \frac{1}{k+1}Sk​=1−k+11​

Step 2: MATLAB Function

You can now write a MATLAB function to compute this sum and plot it. Below is the MATLAB code to calculate the sum for k=1,2,…,1000k = 1, 2, \dots, 1000k=1,2,…,1000 and create a semi-logarithmic plot:

matlabCopyEditfunction semilogx_sum
    % Define the number of terms
    k_max = 1000;
    
    % Pre-allocate the array for storing sums
    S = zeros(1, k_max);
    
    % Compute the sum for each k
    for k = 1:k_max
        S(k) = 1 - 1/(k+1);
    end
    
    % Plot the result in semilog-x scale
    semilogx(1:k_max, S, 'LineWidth', 2);
    title('Semilog-x Plot of the Sum S_k');
    xlabel('k');
    ylabel('S_k');
    grid on;
end

Step 3: Deriving the Infinite Sum

As kkk approaches infinity, the term 1k+1\frac{1}{k+1}k+11​ becomes very small, so the sum SkS_kSk​ approaches:S∞=1−0=1S_{\infty} = 1 – 0 = 1S∞​=1−0=1

Thus, the infinite sum is 1.

Step 4: Plotting Interpretation

When you run the MATLAB function, you’ll notice that the plot levels off as kkk increases, approaching the value of 1, which corresponds to the sum as k→∞k \to \inftyk→∞. The semi-logarithmic plot helps highlight how the sum converges as kkk increases.

Scroll to Top