7

In other words, does the inequality,

$$\int_1^{i+1}f(x) \, dx \leq \sum_{n=1}^i a_n \leq \int_1^n f(x) \, dx+a_1,$$

only hold when $f(x)$ is continuous, monotone and decreasing?

Also $f(n)=a_n$.

RRL
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J.Gudal
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  • It's commonly proved under these assumptions, but I strongly suspect that these assumptions could be weakened a little. For example, I doubt you need continuity, only that the various integrals exist. – Dave L. Renfro Jan 11 '16 at 17:30
  • @DaveL.Renfro I assume that the decreasing condition is necessary right? – J.Gudal Jan 11 '16 at 17:31
  • I doubt the decreasing condition is necessary either. It seems to me that standard "picture proof" with rectangles lying below the curve would still work if the curve had lots of wiggles in which the wiggles were of sufficiently small amplitude that you still have the curve lying between the "over-estimate rectangles" and the "under-estimate rectangles". – Dave L. Renfro Jan 11 '16 at 17:59
  • You can get a counterexample to the left hand or right hand integral test by writing a function with a big spike in between each adjacent pair of integers. For instance it could be piecewise linear spiking up to $3$ exactly in the middle of each $[n,n+1]$ and equal to $1/n^2$ at $n$. Then it is not hard to see that the sum converges but the integral is bigger than $\sum_{n=1}^\infty 1 = \infty$. I think there is no counterexample to the upper and lower sum version of the integral test (which is the same as the left hand or right hand version in the decreasing case). – Ian Jan 11 '16 at 19:15
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    Is there a version of this theorem with f only measurable, non-negative, and $\lim\limits_{x \to \infty} f(x) = 0$? – Jeffrey Rolland Jul 08 '16 at 19:14

1 Answers1

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A theorem of Hardy states that for a nonnegative function $f$ (which is not necessarily monotone), the sum $\sum_{j=1}^nf(j)$ and the integral $\int_0^nf(t)\, dt $ converge or diverge together if $f$ has a continuous derivative satisfying

$$\int_0^{\infty}|f'(t)|dt < \infty.$$

The integral test also holds under weaker conditions where it is sufficient only that the total variation of $f$ on $[0,n]$ is bounded for all $n$.

A proof of Hardy's result is as follows:

Let $\{t\}$ denote the fractional part of $t$. Then

$$\int_0^n \{t\}f'(t) \, dt = \sum_{j=1}^n \int_{j-1}^j(t-j+1)f'(t) \,dt.$$

Integrating by parts,

$$\int_0^n \{t\}f'(t) \, dt = \sum_{j=1}^n \left[(t-j+1)f(t)|_{j-1}^j-\int_{j-1}^jf(t) \,dt \right] \\ = \sum_{j=1}^n f(j)-\int_{0}^nf(t)\,dt.$$

Hence, for all $n$

$$|\sum_{j=1}^n f(j)-\int_{0}^nf(t)\,dt| \leqslant \int_0^n|\{t\}f'(t)| \, dt \leqslant \int_0^{\infty}|f'(t)| \, dt < \infty.$$

From this final inequality, we can make comparisons to show that the integral and sum must converge or diverge together.

RRL
  • 90,707
  • Thanks! This sounds a lot more general. – J.Gudal Jan 11 '16 at 23:19
  • How does the proof proceed under the weaker assumption that the total variations on $[0, n]$ are all uniformly bounded? Using summation by parts or some sort of Stieltjes integration? Or approximation by functions whose derivatives are continuous and in $L^1([0, +\infty)$? – Mishel Skenderi Jul 11 '23 at 19:14
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    @MishelSkenderi: It has been a long time, but I think what I meant was based on the following simple observation: $$\left|\sum_{j=1}^n f(j) - \int_0^n f(x) , dx \right| = \left|\sum_{j=1}^n \int_{j-1}^j [f(j) - f(x)] , dx \right| \leqslant \sum_{j=1}^n \sup {|f(j) - f(x)|: j-1 \leqslant x \leqslant j}$$ – RRL Jul 11 '23 at 20:33
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    The the series and integral converge or diverge together if $\sum_{j=1}^\infty \sup {|f(j) - f(x)|: j-1 \leqslant x \leqslant j} < \infty$. – RRL Jul 11 '23 at 20:35