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Practice Paper 1 Question 20

Evaluate \(\lim\limits_{n\rightarrow\infty} \left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}\right)\). Hint: Graph sketching may help.

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Warm-up Questions

  1. Sketch \(y=\frac{x}{x^2-a^2}\) for various values of \(a\).
  2. Integrate \(y=x\ln x\).
  3. Evaluate \(\lim\limits_{x\to\infty} \big(\ln x-\ln(2x-1)\big).\)

Hints

  • Hint 1
    What does the graph of \(\frac{1}{x}\) look like?
  • Hint 2
    On the above graph, try representing each term of the sum as a (very basic) shape with an area equal to the term's value.
  • Hint 3
    Which 2 continuous functions can you use as upper and lower bounds for the terms of the sum, having this new representation? Hint: one of them you already used.
  • Hint 4
    Consider the graph below. How do you relate the areas underneath the two functions to the sum?
  • Hint 5
    Squeeze theorem?

Solution

Each number \(\frac{1}{k}\) is equal to the area of the rectangle extending to the right of the number, having height \(\frac{1}{k}\) and width 1. By sketching, notice there are two continuous functions bounding these rectangles, one above, i.e. \(\frac{1}{x-1}\), and one below, i.e. \(\frac{1}{x}\), whose integrals will also bound the initial sum under the limit.

Concretely,

\[ \int_{n+1}^{2n+1} \frac{dx}{x} < \sum_{k=n+1}^{2n}\frac{1}{k} < \int_{n+1}^{2n+1}\frac{dx}{x-1}. \] At \(n\rightarrow\infty\), both integrals become \(\ln2,\) and by the squeeze theorem, so must the sum.

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