Why is $\int_1^{\infty}(1/x^2)\,dx$  less than $\sum_{n=1}^{\infty} 1/(n^2)$? 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?
 A: Not if the rectangles (i.e. the terms in the series) stick out above the curve we're integrating:
   
(see the page on the integral test)
A: 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 $I=1$). 
A: 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.
