Infinite series and improper integrals

[leehufford]

Hello,

I've been reviewing some calculus material lately and I just have a couple questions:

1) I've seen infinite series shown graphically as a collection of rectangular elements under a curve representing an approximation of the area under the curve. But the outputs of the infinite series are just one dimensional numbers, so wouldn't lines coming from each natural number extending upward to the functional value of the curve be a more realistic graph? How do we get rectangles when we are summing numbers? We don't have a value for n=1.5 when the sum is from n=1 to n= infinity, so why is there rectangular area at x = 1.5 on the graph? (I know I'm wrong-- I just want to know why I am wrong.)

2) I came to my first question by noticing that the infinite series 1/x² converges to pi²/6 while its associated improper integral converges to 1. I guess I feel like whatever differences exist in these values would be "straightened out" or made trivial because we are sending x to infinity.

Thanks in advance!

-Lee

 

[jbunniii]

Hello,

I've been reviewing some calculus material lately and I just have a couple questions:

1) I've seen infinite series shown graphically as a collection of rectangular elements under a curve representing an approximation of the area under the curve. But the outputs of the infinite series are just one dimensional numbers, so wouldn't lines coming from each natural number extending upward to the functional value of the curve be a more realistic graph? How do we get rectangles when we are summing numbers? We don't have a value for n=1.5 when the sum is from n=1 to n= infinity, so why is there rectangular area at x = 1.5 on the graph? (I know I'm wrong-- I just want to know why I am wrong.)

If we make the width of the rectangle equal to 1 and the height equal to $x_n$, then the area is $x_n$. So the area of the $n$'th rectangle equals the $n$'th term of the series. This is useful because it allows us to express the series as an integral (sum of the areas of the rectangles), which we may then compare to (i.e. bound it above and below by) other integrals whose values we know how to compute.

2) I came to my first question by noticing that the infinite series 1/x² converges to pi²/6 while its associated improper integral converges to 1. I guess I feel like whatever differences exist in these values would be "straightened out" or made trivial because we are sending x to infinity.

Unless you make the rectangles thinner and thinner, the sum of the areas of the rectangles will NOT equal the integral of $1/x^2$. At each value of $n$, there is a discrepancy between the area of $n$'th rectangle (which has width 1 and height $1/n^2$) and $\int_{n}^{n+1}(1/x^2) dx$. There is no reason to expect these discrepancies to cancel out as $n \rightarrow \infty$.

 

[leehufford]

If we make the width of the rectangle equal to 1 and the height equal to $x_n$, then the area is $x_n$. So the area of the $n$'th rectangle equals the $n$'th term of the series. This is useful because it allows us to express the series as an integral (sum of the areas of the rectangles), which we may then compare to (i.e. bound it above and below by) other integrals whose values we know how to compute.


Unless you make the rectangles thinner and thinner, the sum of the areas of the rectangles will NOT equal the integral of $1/x^2$. At each value of $n$, there is a discrepancy between the area of $n$'th rectangle (which has width 1 and height $1/n^2$) and $\int_{n}^{n+1}(1/x^2) dx$. There is no reason to expect these discrepancies to cancel out as $n \rightarrow \infty$.

Thanks! That made perfect sense.

-Lee

 

[micromass]

This is also relevant: http://en.wikipedia.org/wiki/Euler–Maclaurin_formula