$\lim\limits_{n\rightarrow\infty} \left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{2n}\right)$. Hint: Graph sketching may help.
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- Sketch
$y=\frac{x}{x^2-a^2}$for various values of$a$. - Integrate
$y=x\ln x$. - Evaluate
$\lim\limits_{x\to\infty} \big(\ln x-\ln(2x-1)\big).$
- Sketch
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What does the graph of
$\frac{1}{x}$look like? -
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.
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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.
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Consider the graph below. How do you relate the areas underneath the two functions to the sum?
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Squeeze theorem?
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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.