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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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