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Define $L(x)=\int_{1}^x {1\over t} dt $

NOTE: I realize that $L(x)$ is the definition of $ln(x)$, but we aren't allowed to use that. Our professor is walking us through the definition of $ln(x)$ and $e^x$.

Part A: Show that $L({1\over x}) = -L(x)$

I've tried several different substitutions for this and even direct proof, but I'm completely stuck after working/thinking about this for the past several hours. Help is greatly appreciated!

Part B: Using Cauchy Criterion showing that the sequence $$s_n = 1 + {1 \over 2} + {1 \over 3} + ... + {1 \over n}$$ is divergent if $m>n$, show that $L(x)$ tendsto $\infty$ as $x \to \infty$.

Basically, I'm trying to show that if the limit of the sequence converges to some L, then the function of that sequence also converges to the same L. I suspect that I need to show that $l(x)=s_n={1 \over x}$, but I don't know how to formally state this idea.

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A. When you write the definition of $L(1/x)$, what do you get? Is there a simple change of variables you can use to relate the new interval of integration to the interval $[1,x]$? –  GEdgar Dec 12 '12 at 17:18
    
Is Cauchy criterion the fact that convergent sequences are Cauchy, or that the convergence of the series for a monotonic sequence is determined by the $2^n$th terms, more usually called Cauchy condensation test? –  ronno Dec 12 '12 at 18:01

4 Answers 4

1 For $x>0$, $$[L(\frac1x)]^{\prime}=L^{\prime}(\frac1x)(\frac1x)^{\prime}=x\frac{-1}{x^2}=-\frac1x=(-L(x))^{\prime}$$ which implies $L(\frac1x)=-L(x)$

2 The Harmonic series diverges. By the integral test, $$\sum_{n=1}^{N}\frac{1}{n}\le \int_{1}^{N}\frac{1}{x}dx $$ and so it follows, $$\lim_{N\to +\infty}L(N)=+\infty$$

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It seems Nameless didn't read the question... –  GEdgar Dec 12 '12 at 17:21
    
@GEdgar How about now? –  Nameless Dec 12 '12 at 17:27

For part A, I think the substitution $s=1/t$ should see you through.

For part B, think about grouping with powers of 2, or something similar. Then compare the sum to the integral. This might help.

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For the first part, $$L(1/x) = \int_1^{1/x} \dfrac1t dt$$ Let $y=1/t$, then $dt = -\dfrac{dy}{y^2}$. $$L(1/x) = \int_1^x y\left( -\dfrac{dy}{y^2} \right) = - \int_1^x \dfrac{dy}y = - L(x)$$ For the second part, note that for all $n \in \mathbb{N}$, we have that $$s_{2n} - s_n = \dfrac1{n+1} + \cdots + \dfrac1{2n} > \dfrac1{2n} + \dfrac1{2n} + \cdots + \dfrac1{2n} = \dfrac12$$ Hence, by Cauchy criteria the sequence diverges. Now note that $$\int_1^x \dfrac{dt}t > \int_1^2 \dfrac{dt}2 + \int_2^3 \dfrac{dt}3 + \cdots + \int_{\lfloor x \rfloor - 1}^{\lfloor x \rfloor} \dfrac{dt}{\lfloor x \rfloor} = \dfrac12 + \dfrac13 + \cdots + \dfrac1{\lfloor x \rfloor} = s_{\lfloor x \rfloor} - 1$$

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$$L\left(\frac{1}{x}\right):=\int_1^{1/x}\frac{dt}{t}\;\;,\;\;u:=\frac{1}{t}\,\,,\,du=-\frac{1}{t^2}dt=-u^2\,dt\Longrightarrow$$

$$L\left(\frac{1}{x}\right)=\int_1^x-\frac{du}{u^2}\cdot u=-\int^x_1\frac{du}{u}=-L(x)$$

Added Part B: For $\,m>n\,$:

$$|s_m-s_n|=\left|\frac{1}{n+1}+\frac{1}{n+2}+\ldots +\frac{1}{m}\right|\geq \frac{m-n}{m}=1-\frac{n}{m}$$

We thus can keep $\,n\,$ fixed and let $\,m\to\infty\,$, getting that $\,|s_m-s_n|\geq 1\,$ , which means the sequence cannot be Cauchy and thus cannot converge. Since it is formed by a sum of positive elements, this means the sequence diverges to $\infty\,$

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Holy cow, thank you so much! You have no clue what a relief this is! Also, how simple that substitution was! I actually tried tht, but, looking back at my work, I made a simple algebra mistake :( –  ray Dec 12 '12 at 17:43
    
Yes, sometimes simple things just slip away by our side without us seeing them. Constant study and doing exercises do a lot to making things better. You have several answers now, so consider upvoting those that you find useful and, eventually, accept the one that you find the best for you. –  DonAntonio Dec 12 '12 at 17:56

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