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The value of $\displaystyle\int_1^{\infty}\frac{1}{x^2}\,dx$ equals 1.

The series $\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^2}$ equals $\pi^2/6 \approx 1.66493$.

Shouldn't the area under the curve of the function be larger than the sum of the terms of the sequence?

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3 Answers 3

up vote 30 down vote accepted

Not if the rectangles (i.e. the terms in the series) stick out above the curve we're integrating:

   enter image description here

(see the page on the integral test)

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if you flip the rectangles, you get $1+\int_1^{\infty}x^{-2}$ as an upper bound –  yoyo May 8 '11 at 23:43
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Why? Because the function $x\mapsto1/x^2$ to be integrated is nonincreasing. For the same reason, the integral $I$ is greater than the series $S$ minus its first term: $S-1<I<S$ (as a matter of fact, $S=1.66493$ and $I=1$).

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As of Arturo's edit of the question, there's a slight grammatical mismatch... –  The Chaz 2.0 May 9 '11 at 2:07
    
Thanks everyone. –  lumcti May 9 '11 at 10:58
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I will write down a brief "lemma" so to say which I've learned from Piskunov's Calculus. Which generalizes this:

Let $y=f(x)$ be a function. Let $a_n = f(n)$ and $s_n = \displaystyle \sum_{k=1}^n a_n$. Then

$$\int_1^{n+1} f(x) dx < s_{n+1} < a_1+\int_1^{n+1}f(x)dx$$

Then $s_n$ converges only if the integral converges, and

$$\int_1^{\infty} f(x) dx < s < a_1+\int_1^{\infty}f(x)dx$$

In your case you have $a_1 = 1$, thus you have $1 < s < 2$, but the improper integral is smaller than the sum, as you can see from the lemma.

This can be derived by simply inspectioning the partial sums as rectangles of height $f(k)$ and width $1$, and comparing the upper and lower sums and the actual integral to the value of the series. Some graphing is all it takes.

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