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Edit: I've modified the sums and integrals below into convergent sums and integrals, but my questions are still the same - how can I convert sums into integrals legitimately? As far as I know, the sole requirement is for mesh(p) to shrink to zero. But I am unsure about any requirements (if any at all) for the discrete variable $n$ before converting it to a continuous variable $t$. Thanks,

I've done all of the necessary background reading on Riemman sums and integrals, but perhaps posting my naive thoughts would help me understand it a little deeper:

So, would this limit

$$\lim_{x\to 0} \sum_{n=1}^{\infty}\frac {1}{n^2} x $$

...convert to this Riemann integral

$$\int_1^{\infty} \frac{1}{t^2}dt?$$

I am thinking that the factor $x$ would play the role of mesh(p) for some partition, $p$. Since $x \to 0$, this factor converts to the integration factor $dt$.

But does the discrete variable $n$ ... convert to the continuous variable $t$? Or, do we also need $n$ to shrink to $0$, too? So that what I really need is

$$\lim_{x\to 0} \sum_{n=1}^{\infty}\frac {1}{n^2x} x $$

in order for it to be a legitimate Riemman sum that converts to the above Riemann integral.

Any ideas are welcome.

Thanks,

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  • $\begingroup$ Thanks so much @Bungo, I have modified the question now so that the sums and integrals are convergent :-) $\endgroup$ – User001 Dec 19 '15 at 20:16
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Let $f(t) = 1/t^2$ and suppose want to approximate $\int_1^{\infty}f(t) dt$ as a Riemann sum.

Suppose we use rectangles of equal width, say $x$. Then the edges of the rectangles occur at the points $t= 1, 1+x, 1+2x, 1+3x, \ldots$. We can express this concisely as a sequence $t_n = 1+nx$, where $n=0,1,2,\ldots$.

Let us consider two approximations.

Approximation from below

Here, we choose the height of each rectangle as tall as possible so that the rectangle remains entirely below the graph of the function. Since $f$ is decreasing, this means that we choose the height of the rectangle so that its upper right corner touches the function. So, the height of the first rectangle will be $f(t_1)$, the height of the second rectangle will be $f(t_2)$, etc.

Then the sum of the areas of these rectangles is $$\sum_{n=1}^{\infty}f(t_n)x$$ since $f(t_n)$ is the height of the $n$th rectangle, and every rectangle has width $x$. Since $f(t_n) = f(1+nx) = 1/(1+nx)^2$, this becomes $$\sum_{n=1}^{\infty}\frac{x}{(1+nx)^2}$$ and we would expect that as we make the width $x$ of the rectangles small, this should converge to the integral of $f$: $$\lim_{x \to 0}\sum_{n=1}^{\infty}\frac{x}{(1+nx)^2} = \int_1^{\infty}\frac{1}{t^2}dt$$

Approximation from above

In this case, we choose the height of each rectangle as short as possible so that the rectangle remains above the graph of the function. Since $f$ is decreasing, this means that we choose the height of the rectangle so that its upper left corner touches the function. So, the height of the first rectangle will be $f(t_0)$, the height of the second rectangle will be $f(t_1)$, etc.

Then the sum of the areas of these rectangles is $$\sum_{n=0}^{\infty}f(t_n)x$$ Since $f(t_n) = f(1+nx) = 1/(1+nx)^2$, this becomes $$\sum_{n=0}^{\infty}\frac{x}{(1+nx)^2}$$ Notice that the only difference versus approximation from below is that the sum starts at $n=0$ instead of $n=1$.

As before, if we make the width $x$ small, we would expect that in the limit, we should get the integral: $$\lim_{x \to 0}\sum_{n=0}^{\infty}\frac{x}{(1+nx)^2} = \int_1^{\infty}\frac{1}{t^2} dt$$


Now let's check that the above makes sense in terms of convergence. Let us define $L(x)$ and $U(x)$ to be the lower and upper approximations (i.e., the approximations from below and above, respectively). Therefore, $$L(x) = \sum_{n=1}^{\infty}\frac{x}{(1+nx)^2}$$ and $$U(x) = \sum_{n=0}^{\infty}\frac{x}{(1+nx)^2}$$ It's easy to see that for each fixed $x \geq 0$, both sums converge absolutely, so $L(x)$ and $U(x)$ are well defined. Moreover, their difference is simply the $n=0$ term: $$U(x) - L(x) = \frac{x}{(1+0x)^2} = x$$ Therefore, as $x \to 0$, we have $$\lim_{x \to 0}\ (U(x) - L(x)) = \lim_{x \to 0}\ x = 0$$ which shows that the difference between the upper and lower approximations converges to zero as we shrink the rectangle widths to zero. Since the integral $\int_1^{\infty}(1/t^2)dt$ is sandwiched between these two approximations, both $U(x)$ and $L(x)$ converge to $\int_1^{\infty}(1/t^2)dt = 1$ as $x \to 0$.

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  • $\begingroup$ Hi @Bungo, thank you so much for this amazingly clear and pretty write-up. I love your explanation on how the error term converges to zero. And thanks, especially, for putting your write-up in the context of a problem that we were just discussing last night. It couldn't be more perfect practice :-) Have a great night, Bungo! Hope to hear from you soon, $\endgroup$ – User001 Dec 20 '15 at 0:50
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The first expression is the product of a convergent sum to zero, and only this. Those. value of the first expression can not be determined.
The second expression is a divergent integral, its value is $+\infty$.
The connection between them is missing.

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  • $\begingroup$ Hi @YuriNegometyanov, I have modified my question so that the example sums and integrals are convergent. What do you think now? In converting to a Riemann integral, does the discrete variable $n$ from the summation first have to "shrink" to zero, just like how mesh(p) is required to shrink to zero? Also, the integral I have $\int \frac{1}{t^2}$ -- what would be its associated Riemann sum? Would it be $\sum \frac{1}{n^2} \frac{1}{n}$? Thanks, $\endgroup$ – User001 Dec 19 '15 at 20:24
  • $\begingroup$ Now they are 0 and 1, but the connection is missing again. It can appear when $x$ becomes the parameter of series, but it is a zero factor. $\endgroup$ – Yuri Negometyanov Dec 19 '15 at 21:01

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