Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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:

$$a_1 = 1-0$$



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???

share|cite|improve this question
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
up vote 3 down vote accepted

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

share|cite|improve this answer
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.

share|cite|improve this answer
could you explain how the lower bound "similarly" follows? – Gautam Shenoy Feb 5 '13 at 8:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.