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f is a real valued $C^1$ function on [0,$\infty$]. Suppose that $\int_1^{\infty}|f'(x)|dx$ converges. Show that Convergence of $\sum_{n=1}^{\infty}f(n)$ $\iff $ convergence of $\int_1^{\infty}$f(x)dx.

My thought is like this: $\int_1^{\infty}|f'(x)|dx$ converge $\Rightarrow$ $\sum_{n=1}^{\infty}\int_n^{n+1}|f'(x)|dx $ converge $\Rightarrow$ $\lim_{n \to \infty} \int_n^{n+1}|f'(x)|dx = 0$ $\Rightarrow$ $\lim_{x \to \infty}f'(x) = 0$

If $\sum_{n=1}^{\infty}f(n)$ converge, then $\lim_{n \to \infty}f(n) = 0$ $\Rightarrow$ $\forall \epsilon>0$, $\exists n_0 >0$, such that n>$n_0$ $\Rightarrow$ $|f(n)|<\epsilon$.

Then I don't know what to continue.

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    $\begingroup$ use the integral of the derivative to estimate the error in replacing the integral with the sum. Then use this and the convergence of the absolute value of the derivative to show that they converge at the same time $\endgroup$ – mlu Dec 4 '15 at 19:46
  • $\begingroup$ The convergence of $\int_1^{\infty} |f'(x)|dx$ does not imply that $f'(x)\to 0$. It does not even imply that $f'$ is bounded. The comment by mlu is a good hint. $\endgroup$ – DanielWainfleet Dec 5 '15 at 1:41
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My hint:

$$\left|\int_k^{k+1}f(x)\,dx-f(k)\right| \leq \max_{k<x<k+1} |f(x)-f(k)| \leq \int_k^{k+1} |f'(x)|\,dx $$

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  • $\begingroup$ I still don't understand why $max |f(x) - f(k)|$<= $\int_k^{k+1} |f'(x)|dx$. Even though, |f'(x)| < $\epsilon$ only during a small interval, but the series is summation till infinity, which might not gurantee that the summation is bounded $\endgroup$ – Yuan Dec 4 '15 at 23:56
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    $\begingroup$ If $k \le x \le k+1 $ then $| f(k) - f(x) | = | \int_k^x f'(t) dt | \le \int_k^x |f'(t)| dt \le \int_k^{k+1} |f'(t) | dt$ $\endgroup$ – mlu Dec 5 '15 at 0:23
  • $\begingroup$ The difference between the sum and the integral is bounded by the integral of the derivative which is bounded, so they converge, and diverge, at the same time. $\endgroup$ – mlu Dec 5 '15 at 0:26
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Hint (based on mlu's comment). Let $f:[1,\infty)\to[0,\infty)$ be decreasing then

$$0\leq \lim_{n\to\infty}\left(\sum_{k=1}^nf(k)-\int_1^{n+1}f(x)~\mathrm{d} x\right)\leq f(1).$$

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  • $\begingroup$ You are correct if f is decreasing, but how can we make that f is decreasing $\endgroup$ – Yuan Dec 4 '15 at 19:54

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