# Show that monotone sequence is reducing and bounded by zero?

Let $a_n=\displaystyle\sum\limits_{k=1}^n {1\over{k}} - \log n$ for $n\ge1$. Euler's Constant is defined as $y=\lim_{n\to\infty} a_n$. Show that $(a_n)^\infty_{n=1}$ is decreasing and bounded by zero, and so this limit exists

My thought:

When I was trying the first few terms for $a_1, a_2, a_3$, I get:

$$a_1 = 1-0$$

$$a_2={1\over1}+{1\over2}-\log2$$

$$a_3={1\over1}+{1\over2}+{1\over3}-\log3$$

It is NOT decreasing!!! what did I do wrong??

Professor also gave us a hint: Prove that $\dfrac{1}{n+1}\le \log(n+1)-\log n\le \dfrac1n$

we need to use squeeze theorem???

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typo corrected. Thanks!! – Paul Feb 4 '13 at 16:37
Certainly it's decreasing. I get $1-0=1$, and $1 + \frac12 -\log2=0.8068528\ldots$, and $1+\frac12+\frac13-\log3=0.73472\ldots$. – Michael Hardy Feb 4 '13 at 16:37
If you didn't get a decreasing sequence of numbers for these first three, I wonder if maybe you pressed the "log" key rather than the "ln" key on a calculator? – Michael Hardy Feb 4 '13 at 16:50
This problem can be solved without calculus. – user 1618033 Feb 4 '13 at 18:58

Certainly it's decreasing. I get $1-0=1$, and $1 + \frac12 -\log2=0.8068528\ldots$, and $1+\frac12+\frac13-\log3=0.73472\ldots$.

$$\frac{1}{n+1} = \int_n^{n+1}\frac{dx}{n+1} \le \int_n^{n+1}\frac{dx}{x} = \log(n+1)-\log n.$$ $$\frac1n = \int_n^{n+1} \frac{dx}{n} \ge \int_n^{n+1} \frac{dx}{x} = \log(n+1)-\log n.$$

So $$\Big(1+\frac12+\frac13+\cdots+\frac1n-\log n\Big) +\Big(\log n\Big) + \Big(\frac{1}{n+1}-\log(n+1)\Big)$$ $$\le1+\frac12+\frac13+\cdots+\frac1n-\log n.$$

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oppssss, then bounded by zero can also be explained. But how do I show limit exists??? – Paul Feb 4 '13 at 16:40
If a sequence is monotone (either increasing or decreasing) and bounded, then it converges. – Michael Hardy Feb 4 '13 at 16:58

An elementary way to see the inequality is to recall the formal definition of $e^x=1+x+\frac{x^2}{2!}+\ldots$, so using $\log(n+1)-\log(n)=\log(1+1/n)$ and $1+1/n\leq e^{1/n}$ gives the result upon taking logarithms. The lower bound follows similarly.

Using the inequality your professor gave, look at $a_{n+1}-a_n$ and you'll see the sum term becomes very simple.

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could you explain how the lower bound "similarly" follows? – Gautam Shenoy Feb 5 '13 at 8:43